Plus Two

Students can Download Chapter 5 Evolution Questions and Answers, Plus Two Zoology Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Zoology Chapter Wise Questions and Answers Chapter 5 Evolution

Plus Two Zoology Evolution One Mark Questions and Answers

Question 1.
S L Miller’s closed flask contained
(a) CH₄
(b) H₂
(c) NH₃ and H₂ O
(d) All of these
Answer:
(d) All of these

Question 2.
Wings of insects and birds are
(a) analogous
(b) homologous
(c) vestigial
(d) atavism
Answer:
(a) analogous

Question 3.
Convergent evolution is shown by
(a) homologous organ
(b) analogous organ
(c) vestigial organ
(d) All of these
Answer:
(b) analogous organ

Question 4.
Which of the following is the most primitive ancestor of man?
(a) Homo habilis
(b) Homo neanderthalensis
(c) Australopithecus
(d) Ramapithecus punjabicus
Answer:
(d) Ramapithecus punjabicus

Question 5.
Thoms of Bougainvillea and tendrils of Cucurbila are examples of
(a) analogous organs
(b) homologous organs
(c) vestigial organs
(d) retrogressive evolution
Answer:
(b) homologous organs

Question 6.
More than one adaptive radiation appeared to have occurred in an isolated geographical area is called as………..
Answer:
Convergent evolution

Question 7.
Explain Oparin and Haldane Hypothesis.
Answer:
According to Oparin and Haldane, the life could have originated form pre-existing non-living organic molecules.

Question 8.
In a class seminar, Varun says that some scientists believe that the life originated form the spores reached on earth from outer space. Name this theory of origin of life.
Answer:
Theory of panspermia

Plus Two Zoology Evolution Two Mark Questions and Answers

Question 1.
Identify the Experimental set up in Evolution and answerthe following.

1. Name the Experiment.
2. Explain the significance of the given experimental setup.

Answer:
1. Urey and Miller experiment.

2. In laboratory primitive atmosphere is created. In closed flask (Spark discharge apparatus) CH₄, H₂, NH_3, and water vapour is taken and heated at 8000c. S.L. Miller observed that the formation of amino acids. This supports theory of chemical evolution.

Question 2.
Fossils are the written document of evolution.

1. Name the branch of biology which deals with the study of fossils?
2. How can we calculate the age of fossil?

Answer:

1. Paleontology
2. Radioactive – dating

Question 3.
Examine carefully the details given and group similar one in the table. Give suitable title for each column. Analyse the importance of each group in evolution. (Wings of bat, flipper of whale, wings of butterfly, forelimbs of horse, wings of birds, rudementary body hair, wisdom teeth, vermiform appendix of man, patagium of draco).
Answer:

Question 4.
Choose the correct answers. (Convergent evolution, Gene flow, Adaptive radiation, Divergent evolution)

a) Darwin’s finches	(a)
b) Homologous organ	(b)
c)	Migration
d) Analogous of gan	(d)

Answer:

a) Darwin’s finches	Adaptive radiation
b) Homologous organ	Divergent radiation
c) Gene flow	Migration
d) Analogous organ	Convergent evolution

Question 5.

Figure shows white-winged moth and dark-winged moth on a tree trunk.

• in unpolluted area
• in polluted area

a. Which evolutionary theory is depicted by this figure?
b. What is industrial melanism?
c. Can you give any other example for this theory?
Answer:
a. Theory of Natural selection by Darwin

b. Industrial melanism is the development of dark colouration in populations that are exposed to industrial air pollution.

c. – Resistance of Mosquitoes against DDT.
– Resistance of some microbes against some antibiotics.

Question 6.
The two scientists set up their experimental apparatus by using following components.
(CH₄, NH₃, H₂, Condenser, Electrode, Vaccum pumps, boiling chambers, gas chamber, ‘U’shaped tubes)

1. State chemical evolution theory.
2. Who experimentally proved this concept?

Answer:

1. Chemical evolution states that first form of life could have come from pre-existing non-living organic molecules, for formation of diverse organic molecules from inorganic constituents.
2. Urey Miller

Question 7.

1. Evolution is a departure from Hardy Weinberg equilibrium. Comment on this statement.
2. What are the five factors that affect Hardy Weinberg equilibrium?

Answer:
1. Hardy-Weinberg principle says that allele frequencies in a population are stable and constant from generation to generation.

The gene pool remains constant, which is called genetic equillibrium. Disturbance in this equillibrium, ie. change of frequency of alleles in a population results in evolution.

2. Five factors are

• gene migration or gene flow
• genetic drift
• Mutation
• genetic recombination
• Natural selection

Question 8.
Abiotic origin of life is in progress on a planet other than earth. Based on the origin of life on earth what should be the conditions on the planet.
Answer:

• High Temperature
• Presence of gases such as C, H, N
• Anaerobic environment
• Torrential rain
• High energy radiation
• Primordial soup

Question 9.
As you know Lamarckism and Darwinism are two theories explaining the mechanism of evolution.

1. Write the difference between acquired characters of Lamarck and variation theory of Darwin. Specify.
2. How Darwin defined variation as a key to evolution?
3. Give one example for Darwinism and Lamarckism.

Answer:
1. Acquired characters are inherited to the next generation. According to Darwin, only favourable varia¬tion is inherited to the next generation, which leads to speciation (origin of species).

2. According to Darwin variations are of two types, namely, somatic variation (in body) and genetic variation, which is a key to evolution.

3. Lamarckism – Origin of long necked giraffe. Darwinism – Industrial melanism in peppered moth.

Question 10.
The following diagram represents the operation of Natural selection on one trait.

1. Identify the type of selection.
2. What is meant by Natural selection?

Answer:

1. Stabilising selection
2. Organism with favourable variations are selected by nature and those without them are rejected. This is called Natural selection.

Question 11.
Construct a flow chart showing Darwinism using following points. (Over production, Struggle for existence, Variation Natural Selection, Gurvival of the fittest, origin of species).
Answer:

Question 12.
Name the gases filled in the gas chamber to design the condition of primitive earth.
Answer:
Methane, Ammonia, H₂, water vapour

Question 13.

1. Identify the type of evolutionary process shown here.
2. Give one more example for this.

Answer:

1. Adaptive radiation
2. Darwin’s finches

Question 14.
“Darwin’s finches are excellent example for adaptive radiation.’’

1. Name the place where Darwin’s finches found.
2. Explain adaptive radiation.

Answer:

1. Galapago islands
2. Adaptive radiation is a process of evolution of different species in a geographical area starting from a point and literally radiating to other areas of geography.

Question 15.
“Galapagos islands are the living laboratory of evolution”.

1. Who made this statement?
2. Name the small dark birds in these islands.

Answer:

1. Darwin
2. Darwin’s finches

Question 16.
Arrange them in chronological order. Homo sapiens, Dryopithecus, Australopithecus Homo habilis Homo erectus Answer:

• Dryopithecus
• Australopithecus
• Homo habilis
• Homo erectus
• Homo sapiens

Question 17.
Write one word for the following:

1. Single-step large mutation causing speciation.
2. Allelic frequencies remain constant from generation to generation.
3. Evolution of different species in a given geographical area starting from a point and literally radiating to other areas of geography.
4. Change in gene frequencies occur by chance in small population.

Answer:

1. Saltation
2. Hardy-Weinberg equilibrium
3. Adaptive radiation
4. Genetic drift

Question 18.
How do Darwin’s finches illustrate adaptive radiation?
Answer:
Original stock of seed-eating finches migrated to different habitats, adapted to different feeding methods, by altered beak structure, evolved into different types of finches.

Question 19.
Odd man out. Justify your answer.

• Recombination
• Genetic drift
• Gene migration
• Natural selection
• Mutation
• Speciation

Answer:
Speciation
All others are mechanisms for variation that leads to the origin of species (speciation).

Question 20.

1. Identify the evolutionary phenomenon shown? Explain.
2. What is founder effect?

Answer:

1. Adaptive radiation
2. The effect where original drifted population becomes founders is called founder effect.

Question 21.
Prior to industrialization, there were far more white winged moths on trees than melanised moths in England. However, after industrialization, the distribution pattern of these two kinds of moths reversed. What does the above observation indicate? Explain giving reasons.
Answer:
It indicates that predators will spot a moth against a contrasting background.

1. After industrialization, the tree trunks became dark due to deposition of soot and smoke; under such a condition, the white winged moths stood out conspicuously and easily detected by the predators and hence they reduced in number.

2. The dark-winged moths could merge with the black colour and they escaped the predators and hence increased in number.

Question 22.
‘Darwin finches are the example of adaptive radiation’. Justify.
Answer:
The beak modifications in finches occurred due to difference in food gathering and feeding habitats of different Islands

Question 23.
Arrange the following into two categories under the heading.
Analogous organ and homologous organ.
Eyes in Human beings and octopus.
Flippers of penguin and dolphin.
Vertebrate hearts.
Thorns and tendril.
Wings of butterfly and birds
Forelimb of vertebrates
Answer:

Analogous Organ	Homologous Organ
Eyes in Human beings and octopus.	Vertebrate hearts.
Flippers of penguin and dolphin.	Thorns and tendril
Wings of butterfly and birds	Forelimb of vertebrates

Question 24.

1. Explain Hardey-Weinberg equilibrium?
2. Mention the factors affecting Hardey-Weinberg equilibrium.

Answer:
1. Allele frequencies in a population are stable and constant from generation to generation. (p2+2pq+q2 =1), p, q are individual frequencies of alleles.

2. The factors are gene migration, genetic drift, mutation, genetic recombination, and natural selection.

Plus Two Zoology Evolution Three Mark Questions and Answers

Question 1.
During his voyage, Darwin went to Galapago islands and observed an amazing diversity of particularly, some black birds, called Finches. Darwin’s Finches represent a best example for an evolutionary phenomenon.

1. Name the phenomenon.
2. In which feature, they showed diversity?
3. Give another example forthis phenomena.

Answer:

1. Adaptive radiation
2. Showed variety in beak structure based on food habits – from seed-eating to insectivorous, vegetarian, etc.
3. Australian Marsupials.

Question 2.

Diagrammatic representation of Miller’s experiment is shown.
1. Label the parts a,b,c,d.

2. Miller’s experiment provided evidence for the theory of………

3. Sequence of substances appearing during the origin of life would have been.

• Amino acids, ammonia, phosphates, nucleic acid
• Ammonia, amino acids, proteins, nucleic acids
• Nucleotides, amino acids, nucleic acids, enzymes
• Enzymes, amino acids, proteins, nucleic acids.

Answer:
1. a – electrodes
b- mixture of gases (CH₄, NH₃, H₂, H₂O)
c – cold water
d – vacuum pump

2. Organic evolution (chemical evolution)

3. ii) Ammonia, amino acids, proteins, nucleic acids.

Question 3.
A list of human ancestry is given. Pick the correct ancestor from the table.
Neanderthal man
Java man
Australopithecus
Cro-magnon man
Dryopithecus
Homo habilus

1. First man ape
2. The stone tool maker
3. Buried their dead.
4. User of fire for hunting, defence and cooking.
5. Hunter with domesticated dog and did cave paintings.
6. Pre-man

Answer:

1. Australopithecus
2. Homo habilus
3. Neanderthal man
4. Java man
5. Cro-magnon man
6. Dryopithecus

Question 4.
Briefly describe about the evolution of Man.
Answer:
1. Dryopithecus and Ramapithecus: were hairy and walked like gorillas and chimpanzees.

2. Ramapithecus: was more man-like while

3. Dryopithecus: was more ape-like.

4. Australopithecines: They hunted with stone weapons and ate fruit… They did not eat meat.

5. Homo erectus: Their brain capacity is around 900cc. They ate meat.

6. Neanderthal man: with a brain size of 1400cc lived in near east and central Asia

7. Homo sapiens: arose in Africa. The cave art developed about 18,000 years ago and agriculture around 10,000 years back.

Question 5.
Match the following

Answer:

Plus Two Zoology Evolution NCERT Questions and Answers

Question 1.
Attempt giving a clear definition of the term species
Answer:
A species is a group of similar individuals differing from the members of other species which interbred freely and relatively stable. Species is the smallest of classification.

Question 2.
Try to trace various components of human evolution (Hint: brain size and funct skeletal structure, dietary preference, etc.).
Answer:
Changes involved in human evolution are as follows

• Flattening of face
•  Reduction in body hair
• Development of curves in the vertebral column for erect posture.
• Bipedal locomotion, arms are shorter than legs
• Increase in brain size and intelligence
• Start eating cooked food.

Question 3.
List 10 modern-day animals and from internet resources link it to a corresponding fossil.
Answer:
Different birds-Archaeopteryx.

Question 4.
Describe one example of adaptive radiation.
Answer:
Darwin’s finches of Qalapago island exhibiting a variety of beaks.

Question 5.
Can we call human evolution as adaptive radiation?
Answer:
No, because in human evolution, brain size, skeletal structure, dietary preference and social and cultural evolution occured while in adaptive radiation, the origin, basic structure and the devleopment of the organs remain same only morphological changes occurs.

Plus Two Zoology Evolution Multiple Choice Questions and Answers

Question 1.
‘Continuity of germplasm’ theory was given by
(a) Hugo de Vries
(b) Weismann
(c) Darwin
(d) Lamarck
Answer:
(b) Weismann

Question 2.
‘Hot dilute soup’ was given by
(a) Oparin
(b) Haldane
(c) Urey
(d) None of these
Answer:
(b) Haldane

Question 3.
Which one of the following theories was proposed by Weismann?
(a) Law of inheritance
(b) Theory of inheritance of acquired characters
(c) Theory of natural selection
(d) Theory of germplasm
Answer:
(d) Theory of germplasm

Question 4.
Coacervatesare
(a) protobionts having polysaccharide, protein, and H₂O
(b) protein aggregate
(c) protein and lipid aggregates
(d) None of the above
Answer:
(a) protobionts having polysaccharide, protein, and H₂O

Question 5.
The idea that the life originates from pre-existing life is referred as
(a) biogenesis theory
(b) special creation theory
(c) abiogenesis theory
(d) extraterrestrial theory
Answer:
(a) biogenesis theory

Question 6.
Synthesis of amino acids to prove that amino acids were formed in primitive ocean was experimentally proved
(a) Oparin
(b) Haldane
(c) Sydney fox
(d) Stanley Miller
Answer:
(d) Stanley Miller

Question 7.
The abiogenesis occurred about how many billion years ago?
(a) 1.2 billion
(b) 1.5 billion
(c) 2.5 billion
(d) 3.5 billion
Answer:
(d) 3.5 billion

Question 8.
Pasteur and Koch are related to
(a) discovery of nucleic acids (DNA and RNA)
(b) discovery of ultracentrifuge
(c) germ theory of disease
(d) gene splicing
Answer:
(c) germ theory of disease

Question 9.
Wings of insects and birds are
(a) analogous
(b) homologous
(c) vestigial
(d) atavism
Answer:
(a) analogous

Question 10.
Convergent evolution is shown by
(a) homologous organ
(b) analogous organ
(c) vestigial organ
(d) All of these
Answer:
(b) analogous organ

Question 11.
Which one of the following is not a vestigial structure in Homo sapiens?
(a) Third molar
(b) Epiglottis
(c) Plica semilunaris
(d) Pyramidalis muscle
Answer:
(b) Epiglottis

Question 12.
Which of the following presumably possesses a cranial capacity larger than modem man?
(a) Neanderthal man
(b) Peking man
(c) Australopithecus
(d) Cromagnon man
Answer:
(d) Cromagnon man

Question 13.
The gas mixture used by Miller in his experiment comprised
(a) CH₄,C0₂,N₂,H₂O
(b) NH₃,C0₂,H₂O, N₂
(c) CH₂,NH₃,N₂,H₂O
(d) CH₄,NH₃,H₂,H₂O
Answer:
(d) CH₄,NH₃,H₂,H₂O

Question 14.
The Mesozoic era is also called as
(a) the golden age of the amphibians
(b) the golden age of the reptiles
(c) the golden age of the mammals
(d) the golden age of the birds
Answer:
(b) the golden age of the reptiles

Question 15.
Darwin travelled in which of the following ship?
(a) HNS Eagle
(b) D Matrica
(c) H M S Beagle
(d) Titanic
Answer:
(c) H M S Beagle

Question 16.
There are two opposing views about origin of Modem man. According to one view, Homo erectus in Asia were the ancestors of modem man. A study of variations of DNA however suggested African origin of modem man. What kind of observation on DNA variation could suggest this?
(a) Greater variation in Africa than in Asia.
(b) Variation only in Asia and no variation in Africa
(c) Greater variation in Asia than in Africa
(d) Similar variation in Africa and Asia
Answer:
(c) Greater variation in Asia than in Africa

Question 17.
Big bang theory is connected with
(a) origin of earth
(b) origin of universe
(c) origin of moon
(d) origin of life
Answer:
(b) origin of universe

Question 18.
The primitive earth contains
(a) oxygen, methane, carbon dioxide, and ammonia
(b) Water vapour, methane, oxygen, and ammonia
(c) methane, carbon dioxide an ammonia
(d) Water vapour, methane, carbon dioxide, and ammonia
Answer:
(d) Water vapour, methane, carbon dioxide, and ammonia

Question 19.
The first form of life could have come from pre-existing non-living organic molecules like RNA, protein. This was proposed by
(a) Oparin and Haldane
(b) Charles Darwin
(c) Louis Pasteur
(d) Alfred Wallace
Answer:
(a) Alfred Wallace

Question 20.
The spark discharge apparatus that was used by millercontains
(a) CH₄, O₂, H₂, NH₃ and water vapaur
(b) CH₄, O₂, NH₃ and water vapaur
(c) CH₄, O₂, NH_3, and H₂
(d) CH₄, H₂, NH₃ and water vapaur
Answer:
(d) CH₄, H₂, NH₃ and water vapaur

Question 21.
The first non-cellular forms of life could have originated
(a) 3000 mya
(b) 2000 mya
(c) 2bya
(d) 4bya
Answer:
(a) 3000 mya

Question 22.
The type of evolution leads to the formation of analogous organs
(a) convergent evolution
(b) divergent evolution
(c) microevolution
(d) chemical evolution
Answer:
(a) convergent evolution

Question 23.
The white winged moth survived and the dark-coloured moth were picked out by predators before industrial revolution in England. It supports the theory that proposed by
(a) Lamark
(b) Darwin
(c) Hugo De Vries
(d) Oparin
Answer:
(b) Darwin

Question 24.
In Galapogos island the original seed-eating features of finches of birds changed and become insectivorous and vegetarian finches. This process of evolution radiating to other areas of geography and results
(a) more than two types of beaks in finches
(b) all finches of islands subjected to adaptive radiation
(c) all finches of islands subjected to discontinuous variation
(d) adaptive radiation not leads to evolution
Answer:
(b) all finches of islands subjected to adaptive radiation

Question 25.
A colony of bacteria growing on a given medium utilise a feed component but change in the medium composition affect the population and can survive under the new conditions. This is come under the theory of
(a) Lamark
(b) Darwin
(c) Louis Pasteur
(d) Malthus
Answer:
(b) Darwin

Plus Two Zoology Evolution SCERT Sample Questions and Answers

Question 1.
Arrange the following primates in correct sequence of evolution to man.

1. Dryopethicus
2. Homo erectus
3. Homo habilis
4. Neanderthal man (2)

Answer:
Dryopithecus – Homo habilis – Homo erectus – Neanderthal man.

Question 2.
Carefully read the following statement and answer the questions. “A population remain constant and stable in its allele frequency from generation to generation. Such population shows genetic equilibrium’.

1. Name the underlying principle in the above statement. (1)
2. Name any two factors which affect genetic equilibrium. (1)

Answer:

1. Hardey – Weinberg Principle
2. Mutation, Natural Selection Migration, Genetic drift.

Students can Download Chapter 4 Molecular Basis of Inheritance Questions and Answers, Plus Two Zoology Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Zoology Chapter Wise Questions and Answers Chapter 4 Molecular Basis of Inheritance

Plus Two Zoology Molecular Basis of Inheritance One Mark Questions and Answers

Question 1.
The process of copying genetic information from one strand of the DNA into RNA termed as
(a) translation
(b) transamination
(c) replication
(d) transcription
Answer:
(d) transcription

Question 2.
An enzyme that uses viral RNA as template for the synthesis of DNA is
(a) RNA polymerase
(b) reverse transcriptase
(c) viral nuclease
(d) RNA replicase
Answer:
(b) reverse transcriptase

Question 3.
The sequence of structural gene in Lac operon concept is
(a) Lac A, Lac Y, Lac Z
(b) Lac A, Lac Z, Lac Y
(c) Lac Y, Lac Z, Lac A
(d) LacZ, Lac Y, Lac A
Answer:
(d) LacZ, Lac Y, Lac A

Question 4.
During transcription, the nucleotide sequence of the DNA stand that is being coded is ATACG, then the nucleotide sequence in the mRNA would be
(a) TATGC
(b) TCTGG
(c) UAUGC
(d) UATGG
Answer:
(c) UAUGC

Question 5.
Anticodon is base triplet on
(a) mRNA complementary to base sequence on rRNA
(b) mRNA complementary to base sequence on tRNA
(c) tRNA complementary to base sequence on rRNA
(d) tRNA complementary to base sequence on mRNA
Answer:
(d) tRNA complementary to base sequence on mRNA

Question 6.
There is a murder, which shocked the entire town. The dead body contain the blood smear of other person. The police suspected two persons. Name the method used to identify the murderer.
Answer:
DNA finger printing

Question 7.
In DNA molecule a nitrogenous base bonded with pentose sugar molecule through a………Bond.
Answer:
Phosphodiester bond.

Question 8.
Histone proteins are………Charged molecule.
Answer:
Positively Charged

Question 9.
Histone octamer bind to DNA to form………..
Answer:
Nucleosomes

Question 10.
The chromosomes are seen on which stage of the cell cycle.
Answer:
Metaphase

Question 11.
The virus which infects the bacteria are called…………
Answer:
Bacteriophage

Question 12.
Histones are the proteins found associated with eucaryotic DNA. Name the two Amino acids which is found in greater quantity in Histones.
Answer:
Lysines and Arginines

Question 13.
Pick the genetic material in human beings (RNA, mRN.tRNA, DNA)
Answer:
DNA

Question 14.
The DNA synthesis in leading strand is
(a) continuous
(b) not as single strand
(c) discontinuous
(d) as Okazaki fragments
Answer:
(a) Continuous.

Question 15.
Expand the following
mRNA = messenger RNA
tRNA = ……………….
rRNA = Ribosomal RNA
hnRNA = ………………
Answer:
tRNA = transfer RNA
hnRNA = heterogenous nuclear RNA

Question 16.
Pick the initiation codon from the triplets codons given below. AAA, UGA.AUG, GUA
Answer:
AUG

Question 17.
From the following triplet codes find out the stop codon (GAU, UCU, UAG, UGG)
Answer:
UAG

Question 18.
A DNA molecule in which both strands have radioactive thymidine is allowed to duplicate in an environment containing non-radioactive thymidine. What will be the correct number of DNA molecule that contain some radioactive thymidine after 3 duplications?
Answer:
Two DNA molecules

Question 19.
A change in a sequence of DNA occurs so that the mRNA codon reads AUC rather than AUU. Both of these code for the amino acid isoleucine. Argue that this is not a mutation.
Answer:
If one defines a mutation as a change in genetic material resulting in a different phenotypic expression. Hence this is not a mutation.

Question 20.
Histones are the proteins found associated with eucaryotic DNA. Name the two Amino acids which is found in greater quantity in Histones.
Answer:
Lysines and Arginines

Plus Two Zoology Molecular Basis of Inheritance Two Mark Questions and Answers

Question 1.
To solve a dispute of parentage, the court put an order to conduct a test to prove the father of the child. Name the test used. Procedure of the test is given below, complete it.

1. Isolation of DNA.
2. DNA is cut using restriction endonuclease.
3. ……………
4. …………..
5. hybridization using VNTR probe
6. ………….

Answer:
Name of the test is DNA finger printing. The procedure is

1. Isolation of DNA.
2. DNA is cut using restriction endonuclease.
3. Separation of DNA fragments by electrophoresis.
4. Blotting of separated DNA fragments on nitrocellulose.
5. hybridization using VNTR probe
6. Detection of hybridised DNA fragments by autoradiography.

Question 2.
DNA finger printing is a technique in molecular biology. Arrange the following steps in sequence. Blotting of DNA fragment to nitro cellulose. Digestion of DNA by restriction endonuclease. Deletion of DNA by restriction endonuclease. Isolation of DNA, separation of DNA fragments by electrophoresis.
Answer:

1. Isolation of DNA
2. Digestion of DNA by restriction endonuclease.
3. Separation of DNA fragments by electrophoresis.
4. Blotting of DNA fragment to nitrocellulose.
5. Deletion of hybridised DNA by autoradiography.

Question 3.
Match the columns A, B, and C.

Answer:

Question 4.

Process of Transcription in Bacteria is shown. Identify any two differences, found in eukaryotec transcription.
Answer:
1. There are 3 RNA polymerases, namely RNA Polymerase I – Transcribes rRNAs . RNA Polymerase II – Transcribes hn RNA (precursor of mRNA) RNA Polymerase III – TranscribestRNA, 5srRNA, SnRNAs.

2. There are complexities like splicing, capping, tailing, etc. in eukaryotic transcription.

Question 5.
Identify the following and differentiate them.

1. AUG
2. UGA

Answer:

1. AUG – Codes for methionine and it acts as initiator codon.
2. UGA – Does not code for any amino acids and acts as stop codon.

Question 6.
Repetitive sequences in a DNA molecule are of no importance. Do you agree?
Answer:
Repetitive DNA is a small stretch of DNA repeated many times. They show high degree of polymorphism and form the basis of DNA finger printing.

Question 7.

Observe the diagram and identify A & B.
Answer:

1. Histone octamer
2. DNA

Question 8.
Match the following experiments and conclusions with respective worker.

a. Transforming Principle	i) Messelson & Stahl
b. DNA is genetic material	ii) Watson & crick
c. Semi conservative mode of DNA replication	iii) Fredrick Griffth
d. Proof of semi conservative replication	iv) Hershey & chase

Answer:

a. Transforming Principle	Fredrick Griffth
b. DNA is genetic material	Hershey & chase
c. Semi conservative mode of DNA replication	Watson & crick
d. Proof of semi conservative replication	Messelson & Stahl

Question 9.

Name the process (a) and (b).
Answer:
a – Transcription,
b-Translation

Question 10.
In bacteria, three major types of RNAs are needed for the synthesis of protein.

1. List them and mention their functions.
2. What are the three steps in transcription in bacteria

Answer:

1. List them and mention their functions:

• mRNA – It provides the template
• tRNA – It brings amino acids and reads the genetic code.
• rRNA – It plays structural and catalytic role during translation.

2. three steps in transcription in bacteria:

• Initiation
• Elongation
• Termination

Question 11.
State whether the following statements are true or false. If false rewrite them by changing the word understand.

1. Double helical of DNA was proposed by Jacob and Monod.
2. Sugar present in RNA is deoxyribose
3. Introns are the coding sequences of an eukaryote gene.
4. The common method of DNA replication is conservative.

Answer:
All are false.

1. Watson and Crick
2. Ribose
3. Exons
4. Semi conservative

Question 12.
The transcription is usually taking place from DNA to RNA. Can you explain any instance where transcription takes place in reverse order?
Answer:
Yes. In some viruses, it takes place by reverse transcription.

Question 13.
In the medium where E. Coli was growing, lactose was added, which induced the lac operon. Why does lac operon shut down after sometime addition of lactose in the medium?
Answer:
Lac operon is shut down after some times when the added lactose is utilized from the medium. It is be-cause the repressor protein binds to the operator region of the operon and prevent RNA polymerase from transcribing the operon.

Question 14.
While an mRNA strand is being translated in the ribosome subunit, the triplets in sequence were UAC and UAG. One of them codes for tyrosine. What is the significance of the other? Pick out the codon and specify.
Answer:
UAG acts as terminator codon thus leads to the termination of polypeptide chain. It does not specify any amino acid.

Question 15.
What are structural genes? Name the three structural genes present in the lac operon of Escherichia coil?
Answer:
Structural genes are those genes which actually synthesise mRNAs. The lac operon of Escherichia coli contains three structural genes (z, y and a)

Question 16.
In a classroom discussion your classmate says that the RNA is more stable than DNA. Do you agree with it. Explain the advantage of DNA over RNA.
Answer:
No. DNA is more stable
While DNA contains deoxyribose, RNA contains ribose (in deoxyribose there is no hydroxyl group attached to the pentose ring in the 2 position). These hydroxyl groups make RNA less stable than DNA because it is more prone to hydrolysis. DNA have

1. Replication
2. Chemically and structurally stable
3. Obey Mendelian Characters

Question 17.

1. Mention any four goals of Human Genome Project.
2. Name two methodologies involved in it.

Answer:
1. Identify 20000-25000 genes in human DNA., determine 3 billion base pairs, store information as database, improve tool for data analysis, transfer related technologies to other sectors such as industries.

2. Expressed Sequence Tag Sequence Annotation.

Question 18.
Explain the transforming principle and its experiment.
Answer:
Mouse + live S strain = mouse died Mouse + live R strain = mouse alive Mouse + heat-killed S strain = mouse alive Mouse + heat-killed S strain along with R strain = mouse died.

(certain factors from heat-killed S strain transforms non-virulent R strain to become S strain ie, transfer of genetic material)

Question 19.
The regulation of gene expression happened at various levels in eukaryotes. Point out the levels of gene expression.
Answer:

• Transcriptional level
• Processing level
• Transporting mRNA
• Translational level.

Question 20.
Chromatin is a lengthy molecule. How is it compactly packed in nucleus?
Answer:
Nucleosome Histones are organised to form a unit of eight molecules called as histone octamer. The negatively charged DNA is wrapped around the positively charged histone octamerto form a structure called Nucleosome.

Question 21.
Observe the figure and answerthe following.

1. Name the biomolecule
2. Explain the function of this molecule.

Answer:

1. tRNA
2. Activation and transport of amino acid Ribosome

Question 22.
Identify figure and answer the following.

1. Name the experiment.
2. Briefly explain this experiments.

Answer:

1. Meselson and Stahl Experiment
2. They transferred the cells into a medium with normal ¹⁴NH₄Cl and took samples at various definite time intervals as the cells multiplied, and extracted the DNA that remained as double stranded helices.

The DNA that was extracted from the culture one generation after the transfer from ¹⁵N to ¹⁴N medium [E coil divides in 20 minutes] had a hybrid DNA.

DNA extracted from the culture after another generation [that is after 40 minutes, II generation] was composed of equal amounts of this hybrid DNA and of ‘light’ DNA.

Plus Two Zoology Molecular Basis of Inheritance Three Mark Questions and Answers

Question 1.
Both RNA and DNA are considered as genetic materials.

1. What are the main criteria for a molecule to act as genetic material?
2. Why DNA is considered as a better genetic material than RNA?
3. Why RNA viruses mutate and evolve faster than DNA virus?

Answer:

1. Main criteria for a molecule to act as genetic material:

• Able to generate replica
• Chemically and structurally stable
• Provide scope for slow mutation, required for evolution.
• Able to express itself in the form of ‘Mendelian characters’.

2. RNA, being catalytic, is chemically more reactive and less stable than DNA. Therefore DNA is a better genetic material as it is less reactive and structurally more stable.

3. RNA is unstable than DNA, so mutate at a faster rate. Viruses having RNA genome, with shorter life span mutate and evolve faster.

Question 2.
The sequence of nitrogen bases on one of the polynucleotide chains of a DNA strand is given below.

1. Write the complimentary strand of this polynucleotide chain.
2. Write the sequence of mRNA framed from the template strand.
3. Name the process in which mRNA is formed.
4. Name the enzyme that catalyses the formation of mRNA

Answer:

1. TAC GGC GAT TTT
2. UAC GGC GAU UUU
3. Transcription
4. RNA Polymerase

Question 3.
The most widely accepted scheme for replication was proposed by Watson & Crick.

1. Name the scheme of replication.
2. Who given the experimental proof forthis?
3. Name the medium used.

Answer:

1. Semi conservative DNA replication
2. Messelson and Stahl
3. Medium containing ¹⁵NH₄Cl and normal ¹⁴NH₄Cl. CsCl used for centrifugation.

Question 4.

1. Identify the representative diagram.
2. Mention its application and future challenge (anyone).
3. Name the methods involved in it.

Answer:

1. Human genome.project (HGP)

2. Application – They can study all the genes in a genome.
Future challenge – Requires expertise and creativity of thousands of scientists of various disciplines.

3. Name the methods involved in it:

• Expressed Sequence Tags (ESTs)
• SequenceAnnotation

Question 5.
The schematic representation of the Lac operon is given below:

1. Label a, b, c, d.
2. Name the inducer in this operon.
3. Write any two other operon.

Answer:

1. a – P b – z, c – y, d – a
2. Lactose
3. Trp operon, ara operon, etc.

Question 6.
In eukaryotes, RNA polymerase II transcribes precursor of mRNA, called hnRNA.

1. What is the full form of hnRNA?
2. Describe the additional processings takes place in an hnRNA to become mRNA.
3. What is gene splicing?

Answer:
1. Heterogeneous nuclear RNA

2. The additional processings takes place in an hnRNA to become mRNA:

i. Capping:
It is the addition of nucleotide (methyl guanosine triphosphate) at 5¹ – end of hnRNA.

ii. Tailing:
It is the addition of adenylate residues at 3¹ – end.

3. Gene splicing is the removal of non-functional introns and joining of exons in a defined order.

Question 7.

The above pictures are the DNA finger prints of a child and suspected persons as the father of the child.

1. Out of the suspected persons, find the true father of the child.
2. How will you identify the father?
3. What is the principle behind this?

Answer:
1. C may be the true father.

2. As the bands presented is the finger print match between child and person C. It denotes that hereditary materials are more or less common in these two bands. Usually a child comes 50% of hereditary materials from his father. So ‘C’ may be the father of the child.

3. Principle behind it is VNTRs. Southern blotting using variable number of Tandem Repeats (VNTR) as a probe.

Question 8.
One of the salient features of genetic code is codon is triplet and 61 codons code for amino acids and 3 codons are stop codons. Make a list of other salient features of genetic code.
Answer:
1. Genetic code is unambiguous and specific:
A codon always specifies the same aminoacid, it cannot code for more than one amino acid.

2. Genetic code is deoenerate:
Some amino acids are coded by more than one codon.

3. Genetic code is universal:
Each codon of it specifies or codes for the same kind of amino acid in all organisms. (Eg. UUU-phenyl alanine).

4. Genetic code is polarised:
In specifying a particular polypeptide, the genetic code has a definite initiation codon and a terminal codon.

Question 9.

The diagram of a tRNA the adapter molecule is given.

1. Identify the main parts and their role.
2. What you mean by aminoacylation of tRNA?
3. What is the indication of end of translation?

Answer:

1. main parts and their role:

• Anticodon loop – It has been complementary to the code.
• Amino acid acceptor end. It binds to amino acids.

2. Activation of amino acids in presence of ATP and linking to tRNA is called aminoacylation or charging of tRNA.

3. At the end of translation, a release factor binds to the stop codon, to terminate translation and release polypeptide from ribosome.

Question 10.

Diagram of a polynucleotide is given.

1. Identify the 3 components of a nucleotide.
2. Name the purines.
3. Mention the types of linkage present in a polynucleotide.

Answer:

1. 3 components of a nucleotide:

• A nitrogen base
• A pentose sugar (Ribose/ Deoxyribose)
• Phosphate group

2. Adenine and Guanine

3. types of linkage present in a polynucleotide:

• N-glycosidic linkage
• Phosphoester linkage
• Phosphodiester linkage

Question 11.
DNA finger printing is widely used in forensic application. Some terms related to DNA finger printing are given. What are they?

1. Repetitive DNA.
2. VNTR
3. PCR

Answer:

1. A small stretch of DNA repeated many times is called repetitive DNA.
2. Variable Number of Tandem Repeats
3. Polymerase chain reaction – Synthesis of multiple copies of DNA of interest.

Question 12.
Look at the figure above depicting lac operon of E.coli.

1. What could be the series of events when an inducer is pre^nt in the medium in which E.coli is growing?
2. Name the Inducer.
3. Who introduce operon model.

Answer:

1. Inducer induces the operon by inactivating the repressor, allowing RNA polymerase access to promotor and transcription proceeds.
2. Lactose
3. Jacob and Monad

Question 13.
An mRNA strand has a series of codons out of which three are given below:

• AUG
• UUU
• UAG

1. What will these codons be translated into?
2. What are the DNA codons that would have transcribed these RNA codons?

Answer:
1. These codons be translated into:

• AUG signals the synthesis of polypeptide (start signal) and codes for the amino acid methionine.
• UUU codes for phenylalanine.
• UAG do not specify any amino acid and hence is called nonsense codon. It signals the termination of polypeptide chain (stop signal).

2. TACAAAATC.

Question 14.
The quest for finding out the complete DNA sequence of human genome lead to a mega project called HGP.

1. Expand HGP.
2. In which area of biology HGP is closely associated?
3. Mention any four important goals of HGP.

Answer:

1. Human Genome Project (HGP)

2. Bioinformatics

3. The important goals of HGP are

• Identify all the genes (20,000-25,000) in human DNA.
• Determine the sequences of the 3 billion chemical base pairs that make up human DNA.
• Store this information in databases.
• Improve tools for data analysis.
• Transfer related technologies to other sectors, like industries.
• Address the ethical, legal, and social issues (ELSI) that may arise from the project.

Question 15.
Complete the flow chart of the steps involved in DNA fingerprinting.

Answer:

1. Digestion of DNA by restriction endonuclease
2. Blotting of separated DNA fragments to synthetic membranes (nitrocellulose)
3. Hybridisation using VNTR probe.

Question 16.
Identify the figure and answer the following.

Name a, b and c

Answer:
a. Promoter
b. Structural Gene
c. Terminator

Question 17.
Name the process involved in the synthesis of protein from mRNA. Explain the process.
Answer:
The activation of amino acids with ATP and linked to tRNA- a process commonly called as charging of tRNA or aminoacylation of tRNA

bF or initiation, the ribosome binds to the mRNA at the start codon (AUG) that is recognised only by the initiator tRNA.

In elongation, amino acid linked to tRNA. and bind to the codon in mRNA by forming complementary base pairs with the tRNA anticodon.

The ribosome moves from codon to codon along the mRNA. Amino acids are added one by one and translated into Polypeptide.

At the end, a release factor binds to the stop codon, terminating translation and releasing the complete polypeptide from the ribosome.

Question 18.
Observe the figure and complete table 1 with name .of gene and table 2 with enzymes produced by structural gene.

Lable	Name of Gone
A
B
C

Answer:

Structural Genes	Enzymes
z
y
a

Answer:

Label	Name
A	Promoter gene
B	Repressor gene
C	Structural gene

Structural Genes	Enzymes
Z	p galactosidase
y	permease
A	transacetyl ase

Question 19.
In a classroom discussion your classmate says that the RNA is more stable than DNA. Do you agree with it. Explain the advantage of DNA over RNA.
Answer:
No. DNA is more stable
While DNA contains deoxyribose, RNA contains ribose (in deoxyribose there is no hydroxyl group attached to the pentose ring in the 2’ position). These hydroxyl groups make RNA less stable than DNA because it is more prone to hydrolysis. DNA have

1. Replication
2. Chemically and structurally stable
3. Obey Mendelian Characters

Plus Two Zoology Molecular Basis of Inheritance NCERT Questions and Answers

Question 1.
If the sequence of coding strange in a transcription unit is written as follows:
5’-ATGCATGCATGCATGCATGCATGCATGCATGC-3’ Write down the sequence of mRNA.
Answer:
The sequence of template strand will be
3’-TACGTACGTACGTACGTACGTACGTACGTACG-5’

Thus, the sequence of mRNA will be
5’-AUGCAAUGCAAUGCAAUGCAAUGCAAUGC

Question 2.
Describe In the medium where E. Coli was growing, lactose was added, which induced the lac operon. Then why does lac operon shut down after some time after addition of lactose in the medium?
Answer:
Lac operon is shut down after some times when the added lactose is utilised from the medium. It is because the repressor protein binds to the operator region of the operon and prevent RNA polymerase from transcribing the operon.

Question 3.
Explain (in one or two lines) the function of following:

1. Promoter
2. tRNA
3. Exons.

Answer:
1. Promoter:
It acts as an initiation signal which function as recognition centre for RNA – polymerase provided the operator gene is switched on.

2. tRNA:
It acts as an adapter molecule which transfers amino acids to ribosomes for synthesis of polypeptides.

3. Exons:
These are coding sequences or expressed sequences in an eukaryotic gene. The exon sequences appear in mature or processed RNA.

Question 4.
Briefly describe the following:

1. Transcription
2. Polymorphism
3. Translation
4. Bioinformatics

Answer:

1. Transcription is the process of creating a RNA copy of a DNA sequence.
2. When different types of phenotype occur in the same species the situation is called polymorphism.

Plus Two Zoology Molecular Basis of Inheritance Multiple Choice Questions and Answers

Question 1.
Select the two correct statements out of the four (l-IV) given below about Lac operon.
I. Glucose or galactose may bind with the repressor and inactivate it.
II. in the absence of lactose the repressor binds with the operator region
III. The z-gene codes for region
IV. This was elucidated by Francois Jacob and Jacques Monad
The correct statement are:
(a) II and III
(b) I and III
(c) II and IV
(d) I and II
Answer:
(c) II and IV

Question 2.
In the Lac operon system, beta-galactosidase is coded by
(a) a-gene
(b) i-gene
(c) l-gene
(d) z-gene
Answer:
(d) z-gene

Question 3.
Match the codons with their respective amino acids and choose the correct answer.

Answer:
(a) 3 4 1 5 2

Question 4.
What is antisense technology?
(a) A cell displaying a foreign antigen used for synthesis of antigens
(b) Production of somaclonal variants in tissue cultures
(c) When a piece of RNA that is complementary in sequence is used to stop expression of a specific gene
(d) RNA polymerase producing DNA
Answer:
(c) When a piece of RNA that is complementary in sequence is used to stop expression of a specific gene

Question 5.
An enzyme that uses viral RNA as template for the synthesis of DNA is
(a) RNA polymerase
(b) reverse transcriptase
(c) viral nuclease
(d) RNAreplicase
Answer:
(b) reverse transcriptase

Question 6.
During translation initiation in prokaryotes, a GTP molecule is needed in
(a) association of 30S, mRNA with formyl-met tRNA
(b) association of 50S subunit of ribosome with initiation complex
(c) formation of formyl- met- tRNA
(d) binding of 30 subunits of ribosome with mRNA
Answer:
(a) association of 30S, mRNA with formyl-met tRNA

Question 7.
The central dogma of protein synthesis in teminism is
(a) rRNA → DNA → mRNA → Protein
(b) DNA → gRNA → mRNA → Protein
(c) DNA → DNA → mRNA → Protein
(d) mRNA →  gRNA →  DNA →  Protein
Answer:
(a) rRNA → DNA → mRNA→ Protein

Question 8.
Which of the following groups of codons codes for amino acid serine?
(a) CUC, CUC, CUAandCUG
(b) UAU, UAC, UGU, and UGC
(c) UCU, UCC, UCA, and UCG
(d) GUU, GUC, GCU, and GCC
Answer:
(c) UCU, UCC, UCA, and UCG

Question 9.
Jacob and Monod studied lactose metabolism in E.coli and proposed operon concept, applicable for
(a) all prokaryotes
(b) all prokaryotes and some eukaryotes
(c) all prokaryotes and all eukaryotes
(d) all prokaryotes and some protozoans
Answer:
(a) all prokaryotes

Question 10.
The sequence of nitrogen bases in a particulars region of the non-coding stand of a DNA molecule was found to be CAT GTTTAT CGC. What would be the sequence of nitrogen bases in the mRNA that is synthesized the corresponding region of the coding strand in that DNA?
(a) GUACAAAUAGCG
(b) GTA CAAATA GCC
(c) CAU GUU UAU CGC
(d) CAAGAATAU GCC
Answer:
(c) CAU GUU UAU CGC

Question 11.
The process of reverse transcription was brought to light by the work of
(a) Geroge Beadle and Edward Tatum
(b) Garrod
(c) HMTemin and D Baltimore
(d) RW Holley and Grover
(e) Marashall and W Nirenberg
Answer:
(c) HMTemin and D Baltimore

Question 12.
Dr. Khurana and his colleagues synthesized an RNA molecule with repeating sequence of UGN bases (UG UG UG UG UG UG). It produced a tetrapeptide with alternating sequence of cysteine and valine. It proves that codons for cysteine and valine is
(a) UGU and GUU
(b) UGUandGUG
(c) UUG and GGU
(d) GUGandUGU
(e) GUU and UGU
Answer:
(a) UGU and GUU

Question 13.
Degeneracy of genetic codes is due to
(a) functional 61 codons and 20 amino acids
(b) functional 64 codons and 20 amino acids
(c) functional 20 codons and 20 amino acids
(d) functional 20 codons and 61 amino acids
Answer:
(a) functional 61 codons and 20 amino acids

Question 14.
A single amino acid is often coded by more than one triplet code. In most of the cases the first two bases are the same but the third base is different. This feature of the genetic codes is called
(a) universality
(b) non-overlapping and commaless
(c) redundancy and degeneracy
(d) non-ambiguity
Answer:
(c) redundancy and degeneracy

Question 15.
Which of the following pairs of chromosomal mutation are most likely to occur when homologous chromosomes are undergoing synapasis?
(a) Deletion and inversion
(b) Duplication and translocation
(c) Deletion and Duplication
(d) Inversion and translocation
Answer:
(c) Deletion and Duplication

Question 16.
Which of the following is generally used for induced mutagenesis in crop plants?
(a) Alpha particles
(b) X-rays
(c) UN (260nm)
(d) Gamma rays (from cobalt 60)
Answer:
(d) Gamma rays (from cobalt 60)

Question 17.
The most likely reason for the development of resistance against pesticides in insect damaging a crop is
(a) random mutations
(b) genetic recombination
(c) directed mutations
(d) acquired heritable changes
Answer:
(a) random mutations

Question 18.
Double helix model of DNA structure was elucidated through X- ray diffraction studies conducted by Watson and Crick with assistance of
(a) Nirenberg and Ochao
(b) Franklin and Wilkins
(c) Har Gobind Khorana
(d) George Gamov
Answer:
(b) Franklin and Wilkins

Question 19.
The approximate distance between the adjacent base pair is
(a) 0.34 nm
(b) 3.4nm
(c) 340 nm
(d) 034nm
Answer:
(a) 0.34 nm

Question 20.
DNA replication takes place in semi-conservative manner was experimentally proved by
(a) Griffith
(b) Messolson and Stahl
(c) Hershey and Chase
(d) McCarthy, Mecloid, and Avery
Answer:
(b) Messolson and Stahl

Question 21.
Okasaki fragments are
(a) short stretch of DNA
(b) long stretch of DNA
(c) joined DNA fragments by ligase
(d) stretches of DNA in leading strand
Answer:
(a) short stretch of DNA

Question 22.
Messolson and Stahl explained the types of DNA molecule formed in different generations as
(a) light DNA
(b) hybrid DNA
(c) heavy DNA
(d) all the above
Answer:
(d) all the above

Question 23.
The base pairs of DNA double helix is given below. Select the suitable mRNA strand that derived from transcription is
3¹-ATTT C C-5¹
5¹-TAAAGG-3¹
(a) UAAAGG
(b) CUUUCC
(c) GAAAGG
(d) CCUUUC
Answer:
(a) UAAAGG

Question 24.
The number of amino acids formed by the various combination of triplet code is
(a) 61
(b) 20
(c) 64
(d) 16
Answer:
(b) 20

Question 25.
RNA processing occurs in
(a) bacteria
(b) viruses
(c) fungi
(d) vi raids
Answer:
(c) fungi

Question 26.
The amino acid serine is coded by
(a) UGC
(b) AGG
(c) UAG
(d) UCC
Answer:
(d) UCC

Question 27.
The significant aspect of reverse transcription is
(a) the flow information from DNA to RNA
(b) the flow information from RNA to DNA
(c) the flow information from RNA to proteins
(d) both a and c
Answer:
(b) the flow information from RNA to DNA

Plus Two Zoology Molecular Basis of Inheritance SCERT Sample Questions and Answers

Question 1.
Read the following and select the correct statement/statements.
(a) 23 s RNA act as a enzyme in prokaryotes.
(b) In prokaryotes DNA is monocistronic
(c) Francis Crick proposed the Central Dogma of Molecular biology.
(d) In Eukaryotes three types of RNA polymerases are present. (an only, a and b, a and c, a and d) (1)
Answer:
(a) 23 s RNA act as a enzyme in prokaryotes
(d) In Eukaryotes three types of RNA polymerases are present.

Question 2.
Study the mRNA segment given below which is ready to be translated into a polypeptide chain and answer the following questions.

1. What is A and B denotes?
2. Write the triplet codon for A and B (1)

Answer:

1. A – start codon B – stop codon
2. A – AUG B – UAA, UGA, UAG

Question 3.
Carefully read the statement given below and correct the digits in the brackets a and b, if it is wrong.

Answer:

1. 2968 instead of 1968
2. No change

Question 4.
Write the functions of the following

1. 5-methyl Guanosine triphosphate (5^m GPPP)
2. DNA Ligase (2)

Answer:

1. act as cap. In capping an unusual nucleotide (methyl guanosine triphosphate) is added to the 5′- end of hnRNA.
2. Joints DNA fragments during replication

Students can Download Chapter 2 Reproductive Health Questions and Answers, Plus Two Zoology Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Zoology Chapter Wise Questions and Answers Chapter 2 Reproductive Health

Plus Two Zoology Reproductive Health One Mark Questions and Answers

Question 1.
“Saheli” contraceptive oral pill was developed at:
(a) Department of Biotechnology, New Delhi
(b) CDRI, Lucknow
(c) NEERI, Nagpur
(d) Department of Molecular Biology, Hyderabad
Answer:
(b) CDRI, Lucknow

Question 2.
Most accepted method of contraception in India is:
(a) Use of condoms
(b) Taking oral pills
(c) UseoflUDs
(d) Surgical method
Answer:
(c) Use of lUDs

Question 3.
Select the sexually-transmitted disease:
(a) Syphilis
(b) Hepatitis-B
(c) Genital herpes
(d) All of these
Answer:
(c) Genital herpes

Question 4.
In IVF technique, fusion of ovum and sperm occurs in:
(a) Fallopian tube
(b) Uterus
(c) Vagina
(d) Culture medium
Answer:
(d) Culture medium

Question 5.
Genital herpes is a non-curable STP even if detected earlier. Name two other non-curable STDs.
Answer:
Hepatitis-B, AIDS

Question 6.
Identify the relationship between first and second word, find out the fourth one.

1. Termination of pregnancy: MTP
   Insemination into uterus:………..
2. LNG 20: Hormone releasing IUD
   Lippes loop:……….

Answer:

1. IUI
2. Non medicated IUD

Question 7.
Name two barrier methods used in contraception.
Answer:
Condom (male/female) and Cervical cap

Question 8.
Progestasert is a hormone-releasing IUD. Identify another hormone-releasing IUD.
Answer:
LNG 20

Question 9.
Genital herpes is a non-curable STP even if detected earlier. Name two other non-curable STDs.
Answer:
Hepatitis-B, AIDS.

Question 10.
Identify the method in which sperm is directly injected into the ovum for the formation of embryo,
(a) Al
(b) IUT
(c) ICSI
(d) ET
Answer:
(c) ICSI

Question 11.
Certain techniques are used to assist infertile couples to produce children.

1. Name the technique used to assist infertile couples.
2. Give two examples for this type of technique which is used to address male infertility.

Answer:

1. IVF (Test tube baby), Al, ZIFT, GIFT, IUI, ICSI

2. Technique which is used to address male infertility:

• Al
• IUI

Question 12.
A female cannot produce ovum, but can provide suitable environment for fertilization and further development. Suggest the ART which is more suitable for the lady.
Answer:
GIFT

Question 13.
After IVF, the embryo is transferred either to fallopian tube or to uterus. Name the ET method used when the embryo has not reached eight celled stage.
Answer:
ZIFT

Plus Two Zoology Reproductive Health Two Mark Questions and Answers

Question 1.
Syphilis, Gonorrhea are the common STDs. Suggest any two method to prevent the infection of disease.
Answer:

• Avoid sex with unknown partner/ multiple partners
• Always use condoms during coitus.

Question 2.
Do you think that the contraceptive measures should be encouraged. Explain your opinion with reason.
Answer:
Yes, because

1. contraceptives motivate smaller families
2. check population explosion
3. prevent unwanted pregnancies
4. help to space children.

Question 3.
A list of contraceptive methods are given. Categorise them into natural, barrier, IUDS, Oral contraceptives, Surgical. Condoms, Coitus interruptus, Diaphragms, Vasectomy, CopperT, Pills, Lactational Amenorrhea, Loop, Tubectomy, Cervical caps, Cu 7, Periodic Abstinence.
Answer:

Question 4.
A couple having no kids for several years consulted a doctor. After continuous diagnosis, the Doctor advised hormonal treatment for the lady.

1. Name the hormone, which is deficient in the lady.
2. State how this hormone help the lady to become pregnant.

Answer:

1. LH/FSH/Progesterone
2. FSH helps for the growth and maturation of ovarian follicles. LH helps for ovulation. Progesterone maintains pregnancy.

Question 5.
Given below are four methods and their modes of action in achieving contraception. Match the following.

1. Pill – Prevents sperm reaching cervix
2. Condom – Prevents implantation
3. Vasectomy – Prevents ovulation
4. CopperT – Semen contain no sperm

Answer:

1. Pill – Prevents ovulation
2. Condom – Prevents sperm reaching cervix
3. Vasectomy – Semen contain no sperm
4. CopperT – Prevents implantation

Question 6.
Observe the relationship between the first two words and fill up the blanks.

1. HIV – ELISA test; Typhoid – ________
2. Male – Vasectomy; Female -_______

Answer:

1. Widal test
2. Tubectomy

Question 7.

1. How do the oral pills help in birth control?
2. Name common pills use.

Answer:

1. They are used in the form of tablets which contain progesterone and estrogen. Pills have to be taken daily for a period of 21 days starting within the first five days of menstrual cycle. They inhibit ovulation and implantation.

2. Common pills

• Mala D
• Saheli

Question 8.
Is the use of contraceptives justified? Give reasons.
Answer:
Yes, the use is justified.
Population in India is increasing at a very fast rate. To overcome this problem and to have small family, the use of contraceptive could be the best option.

Question 9.
Tubectomy is a method of female sterilization. How is it surgical procedure helps in female sterility?
Answer:
In tubectomy, a small part of the fallopian tube is removed ortied up through a small incision, so that no ovum can be transported to uterus, thus preventing pregnancy.

Question 10.

1. Do you think that reproductive health in our country has improved in the past 50 years?
2. If yes, mention some such areas of improvement.

Answer:

1. Yes, the programmes like family planning and reproductive and child health care have been successfully launched.

2. Now more people are aware about the advantages of small family and are accepting two children norm, some such areas of improvement are massive child immunization, increasing the use of contraceptives, family planning etc.

Question 11.
A mother of one-year-old daughter wanted to space her second child. Her doctor suggested CuT. Explain its contraceptive actions.
Answer:
CuT release Cu²⁺ ions which suppress sperm motility. Increases Phagocytosis of sperms and reduces their fertilizing capacity.

Question 12.

1. Name the hormonal composition of the oral contraceptive used by human females.
2. Explain how does it act as a contraceptive?

Answer:

1. Pills contain progesteron – estrogen combination.
2. They inhibit ovulation and implantation.

Question 13.
A couple who visited an infertility clinic realised that the lady can not produce an ovum but conceive provide suitable environment and deliver a child. Doctor suggested ‘ART’ for them. Suggest the most suitable ART technique for this couple. Write the reason for suggesting the method.
Answer:
GIFT-(Gamete Intra fallopian transfer). Here transfer of an ovum collected from a donor into the fallopian tube of the female, who can not produce ovum.

Or IVF, where Ova from donor and husband’s sperms are induced to form zygote in the laboratory, then transfer into the wife’s uterus/oviduct.

Question 14.
Contraceptive methods are used to block unwanted pregnancies. What are the qualities of an ideal contraceptive?
Answer:
User-friendly, low cost, easily available, reversible, less side effects. Do not interfere sexual act or sexual drive.

Question 15.
Amniocentesis is a prenatal diagnostic test. Now a days it is widely misused as a test for determining the sex of the foetus. The misuse of the test for foetal sex determination is legally banned today. As a plus two biology student give your comments on the issue.
Answer:
Primary objective is prenatal diagnosis of Genetic disorders. However, it used widely for foetal Sex determination leading to female foeticide

Question 16.
The permanent methods of contraception are more effective’ in contraception than any temporary methods.

1. Identify two permanent methods of contraception.
2. Mention the demerits of permanent method.

Answer:

1. Vasectomy and Jubectomy

2. demerits of permanent method:

• Irreversible method
• require hospitalisation

Question 17.
Government of India legalized MTP in 1971 with some strict conditions to avoid its misuse.

1. Define MTP.
2. Suggest the possibilities to legally perform MTP.

Answer:

1. Medical termination of pregnancy

2. Suggest the possibilities to legally perform MTP:

• In conditions of poor maternal health.
• To avoid unintended pregnancies
• unprotected sex or failure of contraceptives lead to unintended pregnancies.

Question 18.
Diseases or infections which are transmitted through sexual intercourse are collectively called STDs.

1. Suggest two methods to prevent STD.
2. Name two examples for STD.

Answer:

1. Suggest two methods to prevent STD:

• Avoid sex with unknown persons
• use condom
• if doubt go to a qualified doctor for early detection.

2. Gonorrhoea, Syphilis, Genital herpes, Chlamydiasis, Genital wart, Trichomoniasis, Hepatitis-B and AIDS.

Question 19.
Population explosion is major problem facing India.

1. Justify the statement with reasons
2. Suggest any two relevant measures to control overpopulation.

Answer:

1. Population explosion is the major single cause behind Poverty and increase competition.
2. Family planning method. Natural method or IUD or use of contraceptives or raising of marriageable age.

Question 20.
AIDS can be transmitted through sexual contact.

1. Expand AIDS.
2. Suggest any two other methods by which humans could be affected with AIDs.

Answer:

1. Acquired Immuno Deficiency Syndrome
2. Blood transfution, Infected mother to foetus or intravenous drug injection.

Question 21.
On clinical examination, it is found that a pregnant lady carrying a defective foetus.

1. Suggest a technique used to get rid of the defective foetus.
2. Name anyone demerit found associated with the suggested technique.

Answer:
1. MTP (Medical termination of pregnancy.
2. Medical Termination of Pregnancy (MTP) is a procedure that is carried out under anaesthesias increases the risk for the procedure. There are high chances of patient having recurrent abortions or may be negatively used to kill female foeticide.

Question 22.
In India, reproductive health care has been improving because of new and effective policies taken up by the health department. Name two programmes launched by Govt, of India to attain reproductive health among the people.
Answer:
Family planning Reproductive & child health care programmes (RCH)

Question 23.
CuT is a contraceptive device.

1. Suggest the contraceptive action of CuT.
2. Name two hormone-releasing lUDs.

Answer:

1. Prevent sperm motility. Or Phagocytosis of sperm
2. Progestasert, LNG 20

Plus Two Zoology Reproductive Health Three Mark Questions and Answers

Question 1.
A newly married couple wants temporary birth control measures for six months.

1. Which contraceptive is ideal for them?
2. What are the features one would find in an ideal contraceptive?
3. What is the advantages and disadvantages of natural methods?

Answer:

1. Condoms/oral contraceptive pills
2. It must be user-friendly, easily available, effective and reversible with nor or lease side effect and in no way interfere with sexual drive.
3. The advantage of natural methods is that there is no side effects. But the chances of failure is high.

Question 2.
Notice shown in front of recognised scanning centres says ‘Sex determination is illegal’.

1. Name the method of sex determination in the early stages other than scanning.
2. What is the advantage of this technique?
3. How people misuse it?

Answer:

1. Amniocentesis.
2. It is a foetal sex determination test based on chromosomal pattern in the amniotic fluid. It helps to know the chromosomal abnormalities in advance.
3. Sex determination sometimes leads to the killing of female foetus.

Question 3.
Match the items in column A with those of column B.

Column A	Column B
a. Lactational amenorrhea	Barrier method
b. Diaphragm	lUD’s
c. Hormone releasing	Oral contraceptives
d. Progesterone	Natural method
e. Tubectomy	Progesterone
f. Injectables	Oviduct

Answer:

Column A	Column B
a. Lactational amenorrhea	Natural method
b. Diaphragm	Barrier method
c. Hormone releasing	lUD’s
d. Progesterone	Oral contraceptives
e. Tubectomy	Oviduct
f. Injectables	Progesterone

Question 4.
A 9-year-old girl accidentally consumes contraceptive pills regularly for 2 months.
(Hint: Contraceptive pills contains estrogen)

1. What type of change you expect in the body of the girl?
2. How it affects her reproductive development?
3. Do you think the stoppage of the pills will reverse the system? If so, give details.

Answer:

1. The girl begins to exhibit female secondary sexual characters.
2. Overall physiological processes involved in reproduction increases.
3. If she attains puberty during this stage, the system will continue, otherwise, it may be reverse the system.

Question 5.
Match the following.

Column A	Column B
a. MTP	Sex determination of foetus
b. STD	Zygote intrafallopian transfer
c. Sterilisation	Medical termination of pregnancy
d. IUD	Syphilis
e. Amniocentesis	Terminal method to prevent pregnancy
f. ZIFT	Copper T

Answer:

Column A	Column B
a. MTP	Medical termination of pregnancy.
b. STD	Syphilis
c. Sterilisation	Terminal method to prevent pregnancy
d. IUD	Copper T
e. Amniocentesis	Sex determination of foetus
f. ZIFT	Zygote intrafallopian transfer

Question 6.
Infertility cases are increasing now a days. These couples could be assisted to have children through certain special techniques called ART.

1. What is ART?
2. Name two methods in ART.
3. Why IVF is called test-tube baby?

Answer:
1. Assisted reproductive technique.

2. two methods in ART:

• GIFT (Gamete Intra Fallopian Transfer)
• ZIFT (Zygote Intra Fallopian Transfer)
• IVF (Invitro fertilization)
• ICSI (Intra Cytoplasmic Sperm Injection)
• IVI (Intra Uterine Insemination)

3. In IVF, ova from the donor and sperm from the male donor are collected and induced to form zygote under stimulated conditions in the laboratory.

Question 7.
The incidences of STDs are high among persons in the age group of 15-24 years:

1. What is STD?
2. Name some common STDs (Any four).
3. What are the principles you have follow to prevent it?

Answer:

1. Sexually transmitted diseases.

2. Gonorrhoea, syphilis, genital herpes, hepatitis B, AIDS, genital warts, trichomoniasis (any four)

3. principles you have follow to prevent it:

• Avoid sex with unknown partners/multiple partners.
• Always use condoms during coitus.
• In case of doubt, go for early detection and get complete treatment if diagnosed with disease.

Question 8.
Different type of contraceptives are available. A couple wants to space children after their first child.

1. What is the ideal contraceptives suggested for female to delay pregnancy or space children?
2. Name the surgical sterilization methods adopted to prevent any more pregnancies.
3. What is the action of Cu ion releasing contraceptives?

Answer:

1. IUDs (Intra Uterine Devices) like CopperT, Lipper Loop, Cu 7 etc.
2. Vasectomy for male and Tubectomy for female.
3. Cu ions released within the uterus suppress sperm motility and the fertilising capacity of sperms.

Question 9.
Nearly 45 to 50 million MTPs are performed in a year all over the world which accounts to 1/5th of the total number of conceived pregnancies in a year.

1. What is MTP?
2. What is the actual aim of it?
3. Which period of pregnancy is safe for MTP?

Answer:
1. Medical Termination of Pregnancy or induced abortion.

2. The aim of MTP is to get rid of unwanted pregnancies either due to casual unprotected intercourse or failure of contraceptives used during coitus. It is essential, when continuation of pregnancy could be harmful or fatal to mother or foetus or both.

3. MTPs are relatively safe during the first trimester.

Question 10.
In India, often females are blamed for the couple being childless.

1. Do you agree with the statement? Why?
2. How a male, who cannot inseminate the female could be corrected?
3. How IUI is conducted?

Answer:

1. There is equal chances for a male or a female partners to cause infertility.
2. Artificial insemination
3. The semen collected from the husband or a healthy donor is artificially introduced into the uterus of female is called Intra Uterine Insemination (IUI).

Question 11.
Match Column A with Column B and Column C.

Answer:

Plus Two Zoology Reproductive Health NCERT Questions and Answers

Question 1.
Suggest the aspect of reproductive health which need to be given special attention in the present scenario.
Answer:
Control on over population through programmes like family planning and Reproductive and child health care (RCH).

Creating awareness among the people about various aspects such as STDs, available birth control methods, postnatal care of mother and child, adolescence and related changes, sex education, etc.

Question 2.
Is sex education necessary in schools? Why?
Answer:
Yes. It will provide right information about sexto the children. Such education will also discourage children from believing in myth and misconception about sex related aspects.

Question 3.
Do you think that reproductive health in our country has improved in the past 50 years? If yes, mention some such areas of improvement.
Answer:
Yes, the programmes like family planning and reproductive and child health care have been successfully launched.

Now more people are aware about the advantages of small family and are accepting two children norm. Some such areas of improvement are massive child immunization, increasing use of contraceptives, family planning etc.

Question 4.
What are the suggested reasons for population explosion?
Answer:

1. Decline in death rate, maternal mortality rate and infant mortality rate.
2. Increase in number of people in reproductive age.
3. Control of diseases
4. Better public health care and greater medical attention.

Question 5.
Is the use of contraceptives justified? Give reasons.
Answer:
Yes, the use is justified. Population in India is increasing at a very fast rate. To overcome this problem and to have small family,

Question 6.
Anniocentesis for sex determination is banned in our country. Is this ban necessary? Comment.
Answer:
In India, Amniocentesis was banned under prenatal Diagnostic Techniques (Regulation and Prevention of Misuse) Act which was enforced in 1994. According to this Act, all the genetic counseling centres and laboratories are required to apply for registration.

It was must because the technique was being misused. Mothers started even get their normal foetus aborted if it was a female. So this Act is aimed to legally check increasing female foeticide which is causing an unfavourable sex ratio.

Question 7.
Suggest some methods to assist infertile couples to have children.
Answer:
Infertile couples may be helped to have children by the following techniques:
Technique

1. In-vitro-fertilization-embryo transfer (IVF-ET) technique:
ZIFT (Zygote intrafallopian transfer); Ovum of female and sperms of male are induced to form zygote in a culture medium and then embryo is transferred in the female for further developments.

It is of two types: Embryo is transferred in the fallopian tube of female at the stage of 8 blastomeres. IUT (Intra-uterine transfer); Embryo is transferred in the uterus at 32-celled stage

2. GIFT (Gamete intrafallopian transfer):
Transfer of an ovum of a female into fallopian tube of another female for fertilization by sperms of former’s husband.

3. ICSI (Intra-cytoplasmic sperm injection):
Sperm of male is directly injected into ovum of female.

4. Al (Artificial insemination):
Sperms of husband (AIH) or a donor male are artificially introduced into the vagina or uterus of the female.

Question 8.
Removal of gonads cannot be considered as a contraceptive option. Why?
Answer:
It is because an ideal contraceptive should be

1. user-friendly
2. with no or least side effect
3. should not interfere with sexual drive, desire and the sexual act of the user.

When gonads are removed, the basic function of reproductive system such as secretion of hormones and gamete production is not possible.

Plus Two Zoology Reproductive Health Multiple Choice Questions and Answers

Question 1.
Rapid decline in population due to high mortality rate is called:
(a) Population density
(b) Population crash
(c) Population explosion
(d) All of these
Answer:
(b) Population crash

Question 2.
Pre-natal defects in the foetus can be detected by:
(a) Laparoscopy
(b) Genetic engineering
(c) MRI
(d) Aminocentesis
Answer:
(d) Aminocentesis

Question 3.
Saheli, a female antifertility pill, is used:
(a) Daily
(b) Weekly
(c) Quarterly
(d) Monthly
Answer:
(b) Weekly

Question 4.
Fertilization of ovum can be prevented by:
(a) Tubal ligation
(b) Vasectomy
(c) Use of 1UDs
(d) All of these
Answer:
(d) All of these

Question 5.
Infant mortality in India is:
(a) 14/1000
(b) 45/1000
(c) 62/1000
(d) 72/1000
Answer:
(d) 72/1000

Question 6.
On which day, world population touched 5th billion?
(a) May 11, 1985
(b) July 11, 1986
(c) May 11, 1987
(d) July 11, 1987
Answer:
(d) July 11, 1987

Question 7.
According to 2001 census, Indian population was:
(a) 684 millions
(b) 844 millions
(c) 1027 millions
(d) 1128 millions
Answer:
(c) 1027 millions

Question 8.
Which period of menstrual cycle is called risky period of conception?
(a) 3^rd to 7^th day
(b) 7^th to 13^th day
(c) 10^th to 17^th day
(d) 15^th to 25^th day
Answer:
(c) 10^th to 17^th day

Question 9.
Test tube baby is a technique where:
(a) Zygote is taicen from ovary, cultured and Implanted in the uterus
(b) Ovum is taken from ovary, then fertilized outside and then implanted in the uterus
(c) Sperm and ovum are fertilized and grow test tube
(d) None of the above
Answer:
(b) Ovum is taken from ovary, then fertilized outside and then implanted in the uterus

Question 10.
Copper-T loop prevents:
(a) Ovulation
(b) Zygote formation
(c) Fertilization
(d) cleavage
Answer:
(c) Fertilization

Question 11.
Intra-uterine devices (lUDs) are used to prevent
(a) Sperm to reach ovum
(b) sperm to reach female
(c) Sperm from leaving testes
(d) all of these
Answer:
(a) Sperm to reach ovum

Question 12.
11^th July is celebrated as:
(a) World AIDS Day
(b) World Environment Day
(c) World population day
(d) World Science and Technology Day
Answer:
(c) world population day

Question 13.
Test tube baby means a baby born when:
(a) It is developed in a test tube
(b) It develops from a non-fertilized egg
(c) The ovum is fertilized externally and there after implanted in uterus
(d) It is developed by tissue culture method
Answer:
(c) The ovum is fertilized externally and there after implanted in uterus

Question 14.
The first case of IVF-ET technique success, was reported by:
(a) Bayliss and Starling Taylor
(b) Robert Steptoe and Gilbert Brown
(c) Louis Joy Brown and Banting Best
(d) Patrick Steptoe and Robert Edwards
Answer:
(d) Patrick Steptoe and Robert Edwards

Plus Two Zoology Reproductive Health SCERT Sample Questions and Answers

Question 1.
Name the surgical methods of birth control, write the merit and demerit of the above methods. (2)
Answer:
In vasectomy, a small part of the vas deferens is removed or tied up through a small incision on the scrotum In tubectomy, a small part of the fallopian tube is removed or tied up through a small incision in the abdomen or through vagina.
Merit – highly effective
Demerit – Reversibility is poor.

Question 2.
In AIDS, HIV attacks on (Helper-T cells, B – Lymphocytes, Monocytes, RBC)    (1)
Answer:
Helper T -Cells

Students can Download Chapter 3 Principles of Inheritance and Variation Questions and Answers, Plus Two Zoology Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Zoology Chapter Wise Questions and Answers Chapter 3 Principles of Inheritance and Variation

Plus Two Zoology Principles of Inheritance and Variation One Mark Questions and Answers

Question 1.
The 1:2:1 ratio with the pink flower in the F, – generation indicate the phenomenon of
(a) dominance
(b) codominance
(c) incomplete dominance
(d) segregation
Answer:
(c) incomplete dominance

Question 2.
Which of these is not a Mendelian disorder?
(a) Turner’s syndrome
(b) Sickle-cell anaemia
(c) Colourblindness
(d) Haemophilia
Answer:
(a) Turner’s syndrome

Question 3.
Crossing over is advantageous because it brings about
(a) variation
(b) linkage
(c) inbreeding
(d) stability
Answer:
(a) variation

Question 4.
Observe the relationship between first two words and suggest a suitable word to the fourth place.

1. Genetic makeup: Genotype: External appearance: …………
2. Multiple alleles: Blood GP: polygenic trait: ………..

Answer:

1. Phenotype
2. Skin colour is human/Kernel colour is wheat.

Question 5.
Law of segregation is also called law of
(a) probability
(b) purity of gametes
(c) independence of gametes
(d) Punnett hypothesis
Answer:
(b) purity of gametes

Question 6.
Complete it.

1. 44+XXY: Klinifetter’s syndrome: 44+XO………….
2. Monohybrid phenotypic ratio: 3:1, Dihybrid phenotypic ratio:………..

Answer:

1. Turner’s syndrome
2. 9:3:3:1

Question 7.

1. 44A + Xo: Turner’s syndrome, 44A+XXY: …………
2. Man: XX-XY: Birds:…………..

Answer:

1. Klinefelter’s syndrome
2. ZW-ZZ

Question 8.
Blood groups of father, mother, and their two children are given below. Work out the genotypes of each blood group.

1. Father-AB group
2. Mother-O group
3. Daughter – A group
4. Son B-group

Answer:

1. Father – l^A l^B
2. Mother – i i
3. Son-l^Bi
4. daughter l^Ai

Question 9.
Why do certain genes tend to be inherited together in a cell at the time of cell division?
Answer:
Due to linkage.

Question 10.
Pedegree analysis is used to study human genetics. Here individuals and their relationship are represented by symbols. Identify the following symbols.

Answer:
(a) Mating between relatives
(b) Female

Question 11.
Select the correctly matched pair/pairs from the following.
(A) Langdon Down-21^st Trisomy
(B) Sutton and Bovery – Principle of inheritance
(C) Henking-X- body (X- chromosome)
(D) Morgan-Mutation
Answer:
A and C

Question 12.
Name the scientists who proposed chromosomal theory of inheritance?
Answer:
Sutton and Bovery

Question 13.
Select the correct statement or statements from the
following.
(A) Gain of additional chromosome is called aneuploidy.
(B) The affected individual of Turner’s and Klinfelter’s syndrome are fertile.
(C) Increase in the whole set of chromosome in an organism is called polyploidy.
(D) Expected F₂ generation ratio of two genes which do not segregate independently is 9:3:3:1
Answer:
A and C are correct

Question 14.
Character of a genetic disorder is given below. ‘Presence of an additional X chromosome making chromosome number 47 (44 autosomes +XXY)’

1. Identify the disorder
2. Write another one character of this disease?

Answer:

1. Klinfelters syndrome
2. Gynacomastia

Plus Two Zoology Principles of Inheritance and Variation Two Mark Questions and Answers

Question 1.
Substitution of wrong amino acid valine instead of glutamic acid at the 6^th position in glob in chain of Hb cause a disease in man.

1. Name the disease.
2. Draw the amino acid sequence Hb^A peptide and Hb^S peptide.

Answer:

1. Sickle cell Anaemia
2. HbA-

Question 2.
Prepare a table chart showing different human blood groups and its possible genotypes.
Answer:

Question 3.
A man had the following symptoms.
A. short stature,
B. round head
C. puffy face
D. mouth always open with uncontrolled salivation, and
E. mental retardation.
Answer:
(a) Down’s syndrome
(b) It is caused by the trisomy of 21^st pair of chromosome, so that there is a total of 47 chromosomes, instead of the normal 46.

Question 4.
Give one example for each of the following.

1. Inborn error in metabolism.
2. Sex-linked recessive disease.
3. 21^st trisomy.
4. Autosome linked recessive disease.

Answer:

1. Phenylketonuria
2. Haemophilia
3. Down’s syndrome
4. Sickle-cell anaemia

Question 5.
Give reason. Phenylketonuria show accumulation of phenyl alanine in blood.
Answer:
In them, phenyl alanine does not convert into tyrosine, due to the lack of enzyme due to gene mutation.

Question 6.
Observe the crossing of Mirabilis plant and the young ones given below.

1. Name the type of phenomenon of the above crossing?
2. Write the ratio in the F2 Generation.

Answer:

1. Incomplete dominance
2. 1:2:1

Question 7.
In an experiment, Red flowered 4 O’clock plant is crossed with a White-flowered 4 O’clock plant. The F1 generation plant is crossed and produced the F2 generation. The phenotypic and genotypic ratio is 1:2:1. Name the type of inheritance? Explain.
Answer:
Incomplete dominance.

Phenotypic & genotypic ratio is 1:2:1.

Question 8.
The percentage of nucleotide Adenine in DNA isolated from human liver is observed to be 30.7%. What is the expected percentage of T, G and C. Justify?
Answer:
T = 30.7%
G = 19.3%
C = 19.3%
The amount of A is always equals to T. So the percentage of T is expected to be very close to 30.7%. G and C together would make up the remainder or 100 – (30.7 + 30.7) = 38.6, thus the percentage of C and G separately would be expected to equal half of 38.6 or 19.3.

Question 9.
Identify the given symbols used in pedigree charts.

Answer:

1. Sex unspecified
2. Parents above and children below.
3. Male

Question 10.
XXY chromosome pattern is the characteristic feature of a genetic disorder.

1. Write the name of the genetic disorder.
2. Write the phenotypic expressions of this disorder.

Answer:

1. Klinefelter’s syndrome
2. Tall stature with feminised characters like gynaecomastia etc. sterile.

Question 11.
Karyotype of human chromosome compliment is given below. It has some abnormality.

1. Identify the abnormality.
2. Mention any one consequence of this abnormality.

Answer:

1. Trisomy in 21^st Chromosome, having a total of 47 chromosomes instead of the normal 46 leading to Down’s syndrome.
2. It leads to physical, psychomotor and mental retardation.

Question 12.
Copy the table and fill the gap appropriately.

Answer:

Question 13.
A man with blood group A married a woman with B group. They have a son with AB blood group and a daughter with blood group O. Workout the cross and show the possibility of such inheritance.
Answer:
Parents – Father – l^A i – A group
Mother -1^A i – B group
Son – l^A l^B – AB group
Daughter – i i – O group

Question 14.
Identify the type of cross made in flow chart. Explain its significance.

Answer:
Test cross
The progenies of test cross can easily be analysed to predict the genotype of the test organism.

Question 15.
If the frequency of a parental form is higher than 25% in a dihybrid test cross, what does that indicate about the two genes Involved?
Answer:
It indicates that linkage is being shown by two genes of parental forms and they are not exhibiting the phenomenon of independent assortment.

Question 16.
How is a child affected if it has grown from the zygote formed by an XX-egg fertilised by a Y-carrying sperm? What do you call this abnormality?
Answer:
The union of an abnormal XX egg and a normal Y- sperm results in trisomy of sex (X) chromosome. This is a type of sex chromosomal abnormality where the individual has 47 chromosomes (44 + XXY) and the abnormality is called Klinefilter’s syndrome.

Question 17.
Who proposed the chromosomal theory of inheritance? Point out any two similarities in the behaviour of chromosomes and genes.
Answer:
Sutton and Boveri

1. Both factors/ genes and chromosomes occur in pairs in normal diploid cells.
2. Both of them segregate/ separate during game to genesis and enter different gametes, i.e., one member of an allelic pair enters one gamete and the other enters another gamete.

Question 18.
If the frequency of a parental form is higher than 25% in a dihybrid test cross, what does that indicate about the two genes involved?
Answer:
It indicates that linkage is being shown by two genes of parental forms and they are not exhibiting the phenomenon of independent assortment.

Question 19.
In a seminar a student says ABO Blood group is an example for multiple alleles. Are you agree with this? Justify your answer.
Answer:
Yes. ABO blood groups are controlled by the gene I, which has three alleles, namely l^A, l^B, i, which produce six different combinations and six genotypes of the human ABO blood types.

Question 20.
Sickle cell anaemia is a genetic disorder. It is produced by gene mutation. Justify the statement.
Answer:
Sickle cell anaemia is caused by the substitution of glutamic acid by valine, which is due to the single base substitution at the sixth codon of the beta-globin gene from GAG to GUG.

Question 21.
Analyse the figure, find out the error and correct.

Answer:

Question 22.
Fill in the vacant columns.

Name of syndrome	Chromosome composition	Symptoms
a	b	Breast develop
c	44AA+xO	d
	Down syndrome	e

Answer:

Name of syndrome	Chromosome composition	Symptoms
Klinefelter’s syndrome	44AA+XXY	Breast develop
Turners syndrome	44AA+XO	Poor breast development
Down syndrome	45A+XX/XY	Short stature Small round head retarded mental development

Question 23.
In an experiment, Red flowered 4 O’clock plant is crossed with a White-flowered 4 O’clock plant. The F1 generation plant is crossed and produced the F2 generation. The phenotypic and genotypic ratio is 1:2:1. Name the type of inheritance? Explain.
Answer:
Incomplete dominance.

Phenotypic & genotypic ratio is 1:2:1.

Question 24.

Find out the eukaryotic ribosome. Justify your answer.
Answer:
Eukaryotic ribosome is 80s ribosome, which is composed of a large sub unit 60s and a small subunits 40s.

Question 25.
A cluster of contrasting traits selected and studied by Mendel is given. Categories them into dominant and recessive traits.

Answer:

Dominant	Recessive
Violet flower	White flower
Green pod	Yellow pod
Axial flower	Terminal flower
Round seeds	Wrinkled seeds

Question 26.
In a hybridization experiment between tall pea plants and dwarf pea plants, a student observed 100% dominant parental traits in F1 generation and 75% dominant, 25% recessive parental traits are observed in F₂ generation.

1. Identify the type of cross
2. Represent F₂ generation using Punnet square.

Answer:
1. Monohybrid Cross
2.

	T	t
T	TT -Tall	Tt- Tall
t	Tt- Tall	Tt- Dwarf

Question 27.
‘Incomplete dominance is an exception to the principle of dominance’

1. Do you agree with statement? Justify?
2. Explain the principle of dominance.

Answer:
1. Yes, in incomplete dominance intermediate character appears and neither character is completely dominant over the other.

2. In law of dominance, in a heterozygous pair of factors one member of the pair dominates (dominant) the other (recessive).

Question 28.
What is test cross? Design a test cross of the following characters/traits of pea plant.

1. Tall X Dwarf
2. Violet flower X White

Answer:
Crossing of F1 progeny with recessive parent

1. Tt × tt, Offsprings 50% Tt & 50% tt
2. Ww × ww, offsprings 50% – Ww & 50% ww

Question 29.
The amino acid composition of a portion of beta polypeptide chain of haemoglobin is given.

1. Identify the beta chain of sickle cell anaemia patient?
2. Write the difference between the Beta chain of normal haemoglobin and the Beta chain of sickle cell anaemia haemoglobin?

Answer:

1. 1 chain
2. Substitution of glutamic acid by valine at the sixth position of beta chain of Hbs peptide.

Question 30.
A representative figure of an individual affected with a chromosomal/genetic disorder is given.

1. Identify the disorder?
2. Write the genetic reason of this disease?

Answer:

1. Down’s syndrome
2. 21^st trisomy/Non-disjunction of chromosomes during Anaphase I of meiosis.

Question 31.
Compare the chromosomal mechanism of sex determination of the following animals.

1. Drosophila
2. Grasshopper

Answer:
1. XX (female) and XY (male) – Male heterogamety – Male has two sex chromosomes X and Y. Half the sperms with X chromosome and the other Half with a Y chromosome.

2. XX (Female) and XO (male) – Male heterogamety – Male has one Sex chromosome, only X chromosome. Half the sperms with X chromosome and the other half without an X chromosome.

Question 32.
ABO blood group is considered as an example of multiple allelism and Codominance. Explain?
Answer:
ABO blood group is controlled by more than two alleles — l^A, l^B and i. A group l^Al^A, l^A,i, B group—l^Bl^B, l^Bi , AB group l^A, l^B, O group – ii

In AB blood group both l^A and l^B alleles behave as dominant genes, producing both A and B antigens.

Question 33.
In Snapdragon F1 progeny shows pink coloured flowers when a cross in made between red flowered and white flowered plants.

1. What is the genetic reason of this phenomenon?
2. Find out the phenotypic ratio of F2 generation of the same cross.

Answer:

1. Incomplete dominance
2. 1 (Red): 2 (Pink): 1 (White)

Question 34.
Categorize following diseases into a sex-linked recessive disorder and autosome linked recessive disorders. Write the characters of each?

1. Haemophilia
2. Phenyl ketonuria

Answer:

1. Sex-linked recessive, absence of blood clotting due to the lack of a protein.
2. Autosome linked recessive, absence of enzymes that convert phenyl alanine to tyrosine.

Question 35.
Differentiate between Homozygous and heterozygous condition with an example.
Answer:
Alleles of a gene are similar in homozygotes,
E.g. TT for tall, tt for dwarf
Alleles of a gene are dissimilar in heterozygote/ heterozygous.
E.g. Tt heterozygous tall.

Question 36.
Analyse the following statements.
(A) Deletions and insertions of base pair of DNA causes Frame shift mutation.
(B) Chemical and physical factors that induce mutation are referred as polyploidy.
Identify the wrong statement and rectify the mistake
Answer:
(B) is wrong
Chemical or physical factors that induce mutation are referred as mutagens and increase in the whole set of chromosome is called polyploidy.

Question 37.
Is it possible for:

1. Awomanto inherit X chromosome from her father?
2. A man to inherit X chromosome from his father?

Answer:

1. Yes. Father give X to female and Y to male offspring
2. No, Father give X to female and Y to male offspring.

Question 38.
Blood groups of father, mother, and their two children are given below. Work out the genotypes of each blood group.

1. Father – AB group
2. Mother – O group
3. Daughter – A group
4. Son B – group

Answer:

1. Father- l^A l^B,
2. Mother- i i,
3. Son – l^Bi,
4. daughter l^A i

Question 39.
‘Down’s syndrome and Turner’s syndrome are the examples of aneuploidy’. Substantiate.
Answer:
Loss or gain of chromosomes is called aneuploidy Down’s syndrome is 21^st trisomy-total number of chromosome is 47, gain of one chromosomes In Turners syndrome one X chromosome is lost which results in 44 autosomes and one sex chromosomes.

Plus Two Zoology Principles of Inheritance and Variation Three Mark Questions and Answers

Question 1.
In one of his dihybrid crosses, Mendel crossed a true breeding double dominant pea plant with round and yellow seeds with true breeding double recessive plant having wrinkled and green seeds. The phenotypic ratio got was 9:3:3:1. Using checker board, try to find out the genotypic ratio.
Answer:

Phenotypic ratio –

1. Round yellow – 9
2. Round green – 3
3. Wrinkled yellow – 3
4. Wrinkled green -1

Phenotypic ratio – 9 3 3 1
Dihybrid genotypic ratio is 1:2:2:4:1:2:1:2:1

Question 2.
State whether the following statements are true or false. If false, rewrite them by changing the word underlined.

1. A chromosome is mostly made up of RNA.
2. Down’s syndrome is caused by the trisomy of the 10^th chromosome.
3. The somatic cells of females have XX pair of sex chromosome.
4. In a normal dihybrid cross, phenotypic ratio is 9:3:3:1.

Answer:

1. A chromosome is mostly made up of DNA.
2. Down’s syndrome is caused by the trisomy of the 21^st chromosome.
3. The somatic cells of females have XX pair of sex chromosome-True.
4. In a normal dihybrid cross, phenotypic ratio is 9:3:3:1 -True.

Question 3.
Chromosomal mechanism of sex determination involves different types of mechanisms. Three such types are given.
XO type
XY type
ZW type

1. Give example for each type.
2. Differentiate between male heterogamety and female heterogamety.
3. Is the sperm or egg responsible for the sex of the chicks?

Answer:
1. XO type – Grasshopper
XY – Man, Drosophila
ZW – Birds

2. Sex determination mechanism in which males produce two different types of gametes (say X and Y) is called male heterogamety and if female produce two types of gametes (Say Z and W), it is called female heterogamety.

3. In chick, it is the egg which determines the sex. ZW is female and ZZ is male. Z and W are produced by the egg not sperms.

Question 4.
Human ABO blood groups are controlled by the gene l, which has three alleles l^A, l^B and i. Each person possesses any two of the three I gene alleles. Make a table showing the different combinations of these three alleles that are possible in human.
Answer:

Question 5.
Experimental verification of the chromosomal theory of inheritance on Drosophila melanogaster led to discovering the basis for variation that sexual reproduction produced.

1. Name the scientist who formulated chromosome theory of inheritance.
2. Who expanded it by experimental verification?
3. Why did he select Drosophila melanogaster for his studies?

Answer:

1. Sutton and Bovery

2. Thomas Hunt Morgan

3. Drosophila was very suitable for his studies because

• They could be grown on simple synthetic medium.
• Complete life cycle in 2 weeks.
• A single mating could produce a large progeny.
• Male and female are easily distinguishable
• Many hereditary variations can be seen with low power microscope

Question 6.
According to common Mendelian pattern of inheritance, F₁ resembles the dominant parent. But in contrast, there are certain other patterns of inheritance in which F₁ does not resembles either of the parents.

1. Name the pattern of inheritance in which F₁ shows a character in between the two parents.
2. Citing an example write the phenotypic ratio of F₂ in the above cross.
3. Name the pattern of inheritance in which F₁ resembles both parents.

Answer:
1. Incomplete dominance

2. Eg: Flower colour in Snapdragon

Phenotypic ratio -1:2:1

3. Co-dominance.

Question 7.

1. What is Pedigree analysis?
2. Write any two Mendelian disorders in man.
3. Give an example each for Autosomal recessive and autosomal dominant traits.

Answer:

1. Analysis of traits in several generations of a family is called pedigree analysis. It helps to study the family history about inheritance.

2. two Mendelian disorders in man:

• Haemophilia
• Sickle-cell Anemia
• Cystic fibrosis
• Colourblindness

3. Autosomal recessive trait – Sickle cell anemia Autosomal dominant trait – Muscular dystrophy

Question 8.
In one of his dihybrid crosses, Mendel crossed a true breeding double dominant pea plant with round and yellow seeds with true breeding double recessive plant having wrinkled and green seeds. The phenotypic ratio got was 9:3:3:1. Using checker board, try to find out the genotypic ratio.
Answer:

1. Round yellow – 9
2. Round green – 3
3. Wrinkled yellow – 3
4. Wrinkled green -1

Phenotypic ratio – 9 3 3 1
Dihybrid genotypic ratio is 1:2:2:4:1:2:1:2:1

Question 9.
What do you understand by the following symbols of Pedigree analysis?

Answer:
(A) Mating
(B) Mating between relatives
(C) Parents above and children (Two) below, one son and one daughter.

Question 10.
‘In human being father is responsible for the sex of child’. Based on |he knowledge of chromosomal mechanism of sex determination, substantiate this statement?
Answer:
During spermatogenesis, two types of gametes are produced of which 50 percent carry the X- chromosome and the rest 50 percent has Y-chromosome besides the autosomes.

Females produce only one type of ovum with an X- chromosome. In case the ovum fertilises with a sperm carrying X- chromosome the zygote develops into a female (XX) and with Y-chromosome results into a male offspring. Thus the genetic makeup of the sperm that determines the sex of the child

Question 11.
Observe the following pedigree chart

1. What is the indication of symbols □ and ○
2. Some symbols are shaded or darkened why?
3. Write the significance of pedigree chart?

Answer:

1. ○ – Female, □ – Male

2. Affected individuals

3. Pedigree charts are important for examining genetics in family. It can be used to track certain traits in family members through two or more generations of a family.

Question 12.
In a pea plant the gene for yellow seed colour (YY) is dominant to green (yy) and round seed (RR) is dominant to wrinkled (rr). With the help of Punnet square, find out the offspring of F₂ generation of the following cross. Yellow Round seeds x Green wrinkled seeds (Homozygous for both characters)
Answer:

Phenotypic ratio: 9 : 3 : 3 : 1.

Question 13.
Match the columns A with column B.

A	B
a. Operon concept	Semiconservative method
b. DNA finger printing	Erythroblastosis foetalis
c. Lathyrus Odaratus	Criss-cross inheritance
d. Rh factor	Lactose-metabolism
e. Haemophilia	Southern Blotting
f. DNA Replication	Linkage

Answer:

A	B
a. Operon concept	Lactose-metabolism
b. DNA finger printing	Southern Blotting
c. Lathyrus Odaratus	Linkage
d. Rh factor	Erythroblastosis foetalis
e. Haemophilia	Criss-cross inheritance
f. DNA Replication	Semiconservative method

Question 14.
Compare incomplete dominance and co dominance with suitable examples?
Answer:

Incomplete dominance	Co dominance
Eg: – Flower color in 4 ‘0’ clock Plant	AB blood group in man
Both dominant and recessive characters are present	Only dominant characters are seen
Dominant character fails to suppress recessive character & Presence of intermediate character	No such suppression. Both dominant characters appear at time.

Plus Two Zoology Principles of Inheritance and Variation NCERT Questions and Answers

Question 1.
Briefly mention the contribution of T.H. Morgan in genetics.
Answer:
Thomas Hunt Morgan put forth chromosome theory of linkage from his work on fruit fly (Drosophila melanogaster). He established the principle of linkage, discovered sex linkage and technique of chromosomal mapping. He wrote a book Theory of gene’ and was awarded Nobel Prize in 1933.

Question 2.
What is pedigree analysis? Suggest how much an analysist can be useful?
Answer:
A record of inheritance of certain traits for two or more generations presented in the form of a diagram or family tree is called pedigree. Analysis of traits in a several generation of a family is called as pedigree analysis. It is employed in case of human beings and domesticated animals.

Importance:
In human genetics, pedigree study provides a strong tool which is utilized to trace the inheritance of a specific trait, abnormality or disease.

It is useful for the genetic counselors to advice in tending couples about the possibility of having children with genetic defects like hemophilia, colour blindness, alkaptonuria, thalassemia, and sickle all anemia.

Question 3.
How is sex determined in human beings?
Answer:
If the zygote is having XX chromosomes in last pair, then it will be female child. If zygote is having XY chromosome then it will be a male child.

Question 4.
A child has blood group O. If the father has blood group A and mother blood group B, work out the genotypes of the parents and the possible genotypes of the other offsprings.
Answer:
The parents of the child will be heterozygous for their blood groups. Therefore, his father with blood group A must-have genotype (l^A l^o) and mother with blood group B must have genotype (l^B l^o).

Question 5.
Explain the following terms with examples:

1. Co-dominance
2. Incomplete dominance

Answer:
1. The alleles which are able to express themselves independently when present together are called co-dominant alleles and this phenomenon is called codominance e.g. AB blood group inheritance in humans.

2. In this phenomenon of the contrasting characters are dominant. The expression of the trait in F1 hybrid is intermediate e.g. pink flowers in peaplant.

Mention the advantages of selecting pea plant for experiment by Mendel. Pea is a annual plant which gives result within a year.

Large number of are produced by pea plant in one generation. Pea plant has short life cy large number of true breeding varities with observable alternative forms trait were available.

Plus Two Zoology Principles of Inheritance and Variation Multiple Choice Questions and Answers

Question 1.
Mating of an organism to a double recessive in order to determine whether it is homozygous or heterozygous for a character under consideration is called
(a) reciprocal cross
(b) test cross
(c) dihybrid cross
(d) back cross
Answer:
(b) test cross

Question 2.
A dihybrid for the traits is crossed with homozygous recessive individual of its type, the phenotypic ratio is
(a) 1 : 2: 1
(b) 3
(c) 1 : 1:1:1
(d) 9:3:3:l
Answer:
(c) 1 : 1:1:1

Question 3.
A dihybrid test cross yielding a result of 1:1:1 ratio is indicative of
(a) four different types of gametes produced by the F1 hybrid
(b) homozygous condition of the F1 -dihybrid
(c) four different types of F1 -generation dihybrids
(d) tour different types of gametes produced by the-parent
Answer:
(a) four different types of gametes produced by the F1 hybrid

Question 4.
When a dihybrid cross is fit into a Punnett square with 16 boxes, the maximum number of different phenotypes available are
(a) 8
(b) 4
(c) 2
(d) 16
Answer:
(b) 4

Question 5.
Cross between unrelated group of organisms is called
(a) hybrid
(b) test cross
(c) hack cross
(d) heterosis
Answer:
(a) hybrid

Question 6.
A pure tall and a pure dwarf plant were crossed to produced offsprings. Offsprings were self crossed, then find out the ratio between true breeding tall to true breeding dwarf?
(a) 1: 1
(b) 3:1
(c) 2:1
(d) 1:2:1
Answer:
(a) 1:1

Question 7.
Genotypic ratio of a monohybrid cross is
(a) 1:2:1
(b) 3:1
(b) 9:3:3:1
(d) !:1:1
Answer:
(a) 1:2:1

Question 8.
Ratio of progeny, when a red coloured heterozygote is crossed with a white coloured plant in which red colour is dominant to white colour
(a) 3:1
(b) 1:1
(c) 1:2:1
(d) 9:3:3:1
Answer:
(b) 1:1

Question 9.
Leaf colour in Mirabilis jalapa is an example of
(a) non-Mendelian inheritance
(b) Mendelian inheritance
(c) chemical inheritance
(d) Both (b) and (c)
Answer:
(a) non-Mendelian inheritance

Question 10.
The F2-generation offspring in a plant showing incomplete dominance, exhibit
(a) variable genotypic and phenotypic ratio
(b) a genotypic ratio of 1:1
(c) a phenotypic ratio of 3: 1
(d) similar phenotypic and genotypic ratio of 1:2:1
Answer:
(d) similar phenotypic and genotypic ratio of 1:2:1

Question 11.
The total number of progeny obtained from a dihybrid cross is 1280 in F2 -generation. How many of them are recombinant type?
(a) 240
(b) 360
(c) 400
(d) 720
Answer:
(c) 400

Question 12.
A test cross is performed
(a) by selfing of F7-generation plants
(b) by selfing of F1 -generation plants
(c) to determine whether F1 -plant is homoygous or heterozygous
(d) between a homozygous dominant and homozygous recessive plant
Answer:
(c) to determine whether F1 -plant is homoygous or heterozygous

Question 13.
Crossing over is advantageous because it brings about
(a) variation
(b) linkage
(c) inbreeding
(d) stability
Answer:
(a) variation

Question 14.
If a plant having yellow or round seeds was crossed with another plant having green and wrinkled seeds then F1 -progeny are in the ratio
(a) 15: 1
(b) 1 15
(c) 1: 13
(d) All yellow and round seeds
Answer:
(d) All yellow and round seeds

Question 15.
When a diploid female plant is crossed with a tetraploid male, the ploidy of endosperm cells in the resulting seed
(a) tetraploidy
(b) pentaploidy
(c) diploidy
(d) triploidy
Answer:
(a) tetraploidy

Question 16.
In certain plant species, red flower colour is incompletely dominant to white flower colour (the heterozygote is pink) and tall stems are completely dominantto dwarf stern. If a tall pink plant (TtRr) is crossed with a tall white plant (TTrr). which one of the following type of plants would be produced in the offsprings
(a) Tall pink and tall white
(b) Dwarf pink and tall red
(c) Dwarf red and tall pink
(d) Tall pink and dwarf white
Answer:
(d) Tall pink and dwarf white

Question 17.
The important events of segregation takes place during
(a) prophase I
(b) anaphase I
(c) prophase II
(d) anaphase II
Answer:
(b) anaphase I

Question 18.
The phenotypic ratio obtained by Mendel in dihybrid cross is
(a) 3:3:1:1
(b) 1:2:1
(c) 9:3:31
(d) 3:1
Answer:
(c) 9:3:31

Question 19.
The unknown genotype of individual is determined by
(a) dihybrid cross
(b) back cross
(c) test cross
(d) monohybrid cross
Answer:
(c) test cross

Question 20.
The studies of incomplete dominance is conducted on
(a) pisum
(b) snapdragon
(c) lathyrus
(d) fruit flies
Answer:
(b) snapdragon

Question 21.
Codominance is shown by the alleles of
(a) l^Ai
(b) l^B i
(c) l^Al^B
(d) ii
Answer:
(c) l^Al^B

Question 22.
The experimental proof for the theory of inheritance was provided by the studies on
(a) pisum
(b) oenothera
(c) lathyrus
(d) fruit flies
Answer:
(d) fruit flies

Question 23.
Gynaecomastia is the case arises by
(a) increase in the number of sex chromosomes
(b) increase in the number of autosomes
(c) increase in the number of autosomes and sex chromosomes
(d) none of the above
Answer:
(a) increase in the number of sex chromosomes

Question 24.
2n-1 is an example of
(a) polyploidy
(b) aneuploidy
(c) euploidy
(d) diploidy
Answer:
(b) aneuploidy

Question 25.
The disease transmitted from father to grandson through daughter in zig-zag manner is an example of
(a) sickle cell anemia
(b) hemophilia
(c) phenyl ketonuria
(d) Turners syndrome
Answer:
(b) hemophilia

Question 26.
Monosomy is represented by the condition of
(a) 2n+1
(b) 2n-1
(c) 2n-2
(d) n-1
Answer:
(b) 2n-1

Plus Two Zoology Principles of Inheritance and Variation SCERT Sample Questions and Answers

Question 1.
A. Carefully read the following symptoms and identify the genetic disorders.

1. Autosomal linked recessive trait where RBC become sickle shaped.
2. Inborn error of metabolism, where phenyl alanine accumulates in blood.
3. Disorderdueto absence of ‘X’ chromosome. (3)

Answer:

1. Sickle cell anaemia
2. Phenyl Ketonuria
3. Turners’ Syndrome

OR

B. Write the contribution of the following biologists in connection with genetics.

1. Henking
2. T.H. Morgan
3. Sutton and Boveri (3)

Answer:
1- Henking- He gave the importance of X body transfer in spermatogenesis of insects

2 – T.H. Morgan – Morgan gave the importance of physical association or linkage of the two genes and coined the term linkage.

3 – Sutton & Boveri – He gave the importance of the pairing and separation of a pair of chromosomes would lead to the segregation of a pair of factors they carried, it is called chromosomal theory of inheritance.

Question 2.
Write the gametes produced from self crossing of genotype RrYy. (2)
Answer:
RY, Ry, rY, ry

Students can Download Chapter 11 Alcohols, Phenols and Ethers Questions and Answers, Plus Two Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Chemistry Chapter Wise Questions and Answers Chapter 11 Alcohols, Phenols and Ethers

Plus Two Chemistry Alcohols, Phenols and Ethers One Mark Questions and Answers

Question 1.
Which of the following is most acidic?
(a) H₂0
(b) CH₃OH
(c) C₂H₅OH
(d) CH₃CH₂CH₂OH
Answer:
(a) H₂O

Question 2.
Propan-1 -ol on reaction with conc.H₂SO₄ at 413 K gives _____________
Answer:
1 -Propoxy propane

Question 3.
Which of the following alcohols can be obtained from HCHO?
(a) CH₃OH
(b) C₂H₅OH
(c) CH₃CH₂CH₂CH₂OH
(d) All of these
Answer:
(d) All of these

Question 4.
Phenol can be distinguished from ethyl alcohol by all reagents except
(a) NaOH
(b) Na
(c) Br₂/H₂O
(d) FeCl₃
Answer:
(b) Na

Question 5.
Anisole on reaction with HI forms
(a) phenol and methyl iodide
(b) iodobenzene and methanol
(c) benzene and methyl iodide
(d) phenol and methanal
Answer:
(a) phenol and methyl iodide

Question 6.
Arrange the following compounds in the increasing order of acidity: Phenol, Alcohol, and Water.
Answer:
Alcohol < Water < Phenol

Question 7.
Phenol is distinguished from ethanol by the following reagents except.
(a) Iron
(b) Sodium
(c) Bromin
(d) NaOH
Answer:
(b) Sodium

Question 8.
Phenol can be converted to o-hydroxy benzaldehyde by
Answer:
Reimer-Tiemann reaction.

Question 9.
4-methoxy acetophenone can be prepared from anisol by ______________
Answer:
Friedel crafts reaction.

Question 10.
Which of the following does not answer idoform test?
(a) 1-propanol
(b) Ethanol
(c) 2 propanol
(d) Ethanal
Answer:
(a) 1- propanol

Question 11.
Chlorination of toluene in presence of light and heat followed by treatment with aqeous KOH gives ________
Answer:
Benzyl alcohol.

Plus Two Chemistry Alcohols, Phenols and Ethers Two Mark Questions and Answers

Question 1.
Reaction of alcohol with metallic sodium is used as a test for alcohol. Substantiate this statement with the help of an equation.
Answer:
Alcohol reacts with metallic Na to form sodium alkoxide with the liberation of H₂.
ROH + Na → RONa + 1/2 H₂

Question 2.
Ethyl alcohol gives iodoform test. Methyl alcohol does not give iodoform test.

1. Do you agree with this?
2. If yes or no, substantiate your view.
3. How can you distinguish between 1 -butanol and 2-butanol?

Answer:
1. Yes.

2. Compounds containing CH₃CO- group and CH₃CH(OH)- group on reaction with iodine and alkali give yellow colour of iodoform. Ethanol contains CH₃CH (OH) – group.

3. 2-Butanol gives iodoform test as it contains CH₃CH(OH)-group, whereas 1-Butanol does not answer iodoform test.

Question 3.
When an organic compound is treated with neutral ferric chloride a violet colour is obtained. What will be the compound? Explain.
Answer:
Phenol. Phenol forms a violet-coloured water soluble complex with ferric chloride.

Question 4.
– O – is the functional group of ether. Classify the following into two groups with appropriate heading.
CH₃-O-CH₃, CH₃-O-C₂H₅, C₂H₅-O-C₂H₅, C₆H₅– O – CH₃.
Answer:

Question 5.
Explain how does the -OH group attached to a carbon of benzene ring activates it towards electrophilic substitution reaction?
Answer:
In phenol, the -OH group is directly attached to a carbon of benzene ring. The lone pair of oxygen participates into resonance with the benzene ring. As a result, electron density on benzene ring increases making it more easy to attack by an electrophile.

Question 6.
Write down the equations for the following conversions using Grignard reagent?
Methanal → Ethanol
Ethanal → 2-Propanol
Answer:

Question 7.
Explain why propanol has higher boiling point than that of the hydrocarbon, butane?
Answer:
Propanol has higher boiling point than butane because it has stronger interparticle forces. In propanol intermolecular hydrogen bonding is present whereas in butane intermolecular forces are weak van der Waals’ forces. A lot of heat is required to break intermolecular hydrogen bonding among propanol molecules.

Question 8.
While separating a mixture of ortho and para nitrophenols by stream distillation, name the isomer which will be steam volatile. Give reason.
Answer:
Ortho-Nitrophenol is steam volatile because in it there is intramolecular hydrogen bonding.

Due to intramolecular hydrogen bonding, the intermolecular forces in ortho-nitrophenol are weaker than that in para-nitrophenol (which has intermolecular hydrogen bonding) and hence it undergoes less association.

Question 9.
Give reason for higher boiling point of ethanol in comparison to methoxy methane.
Answer:
In ethanol the intermolecular forces are hydrogen bonds whereas in methoxymethane the intermolecular forces are dipole-diple forces. Since the intermolecular forces in ethanol are stronger than those in methoxymethane it has higher boiling point than methoxymethane.

Question 10.
While separating a mixture of ortho and para nitro phenols by steam distillation, name the isomer, which will be steam volatile. Give reason.
Answer:
Ortho-Nitro phenol exhibit intramolecular hydrogen bonding and para-Nitro phenol exhibit intermolecular hydrogen bonding. The ortho isomer is steam volatile because there is not intermolecular association.

Plus Two Chemistry Alcohols, Phenols and Ethers Three Mark Questions and Answers

Question 1.
Write a notes on:

1. Rectified Spirit
2. Power Alcohol
3. Denatured Spirit

Answer:
1. 95.6% ethyl alcohol is known as a rectified spirit.

2. A mixture of ethyl alcohol and gasoline can be used as a fuel in the internal combustion engine. It is known as power alcohol.

3. Commercial alcohol is made unfit for drinking by adding certain substances like pyridine, methanol etc. Spirit thus obtained is called denatured spirit.

Question 2.
Fill in the blanks:

Answer:

Question 3.
Ethyl alcohol can be prepared by fermentation of molasses.

1. What do you mean by fermentation?
2. Explain the process.
3. What do you mean by ‘wash’?

Answer:
1. Fermentation is the process of breaking up of large molecules into small molecule in the presence of biological catalyst called enzymes.

2. Molasses is the mother liquor left behind after the crystallisation of cane sugar from sugar cane juice. It is 40% sucrose solution. First it is diluted to 10% solution. Then yeast is added. Temperature of the system is kept at 305 K. The following reactions will take place.

3. 8-10% ethyl alcohol is known as ‘wash’.

Question 4.
Raju prepared soap in the chemistry lab. A liquid remained after the preparation. He argued that it was useless.

1. Do you agree with him? Why?
2. How can you prepare glycerol from spentlye?

Answer:

1. No. It can be used to prepare glycerol.
2. Spentlye contains unreacted alkali, glycerol, water, NaCI, and soluble soap. It is first treated with acid to remove alkali, then with aluminium sulphate to remove soluble soap. It is then evaporated under reduced pressure to remove NaCI.

The resulting solution is then decolorised using animal charcoal, Then the solution is distilled under reduced pressure to remove water. In this way we get glycerol.

Question 5.
When an old sample of ether was heated, it exploded.

1. What is the reason for this phenomenon?
2. How can you detect peroxide content in ether?
3. How can we remove peroxide from old sample of ether?

Answer:

1. Due to the formation of peroxide.
2. Presence of peroxide can be tested by adding ferrous salt solution followed by addition of KCNS solution. Formation of blood red colour indicates the presence of peroxide.
3. The peroxide can be removed by washing with ferrous salt solution.

Question 6.
Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
Answer:
Because of the presence of O-H group in them, alcohols are capable of forming hydrogen bonds with water molecules whereas hydrocarbons cannot form hydrogen bonds with water. As a result, alcohols are more soluble in water than the hydrocarbons of comparable molecular masses.

Question 7.
Give the structures and IUPAC names of monohydric phenols of molecular formula, C₇H₈O.
Answer:

Question 8.
Explain why ortho-nitrophenol is more acidic than ortho-methoxyphenol?
Answer:
ortho-Nitrophenol is more acidic than ortho- methoxyphenol because nitro group by its electron withdrawing resonance effect stabilises the phenoxide ion whereas methoxy group by its electron releasing effect destabilises the phenoxide ion. Greater the stability of the phenoxide ion, greater is the dissociation of phenol and greater is its acid strength.

Question 9.
With the help of a mixture of con. HCI and ZnCI2 how can you distinguish between 1°, 2°, 3° alcohols.
Answer:

• 1° alcohol + Lucas reagent → no reaction
• 2° alcohol + Lucas reagent → turbidity within five minutes
• 3° alcohol + Lucas reagent → turbidity occurs immediately

Lucas reagent anhydrous ZnCI₂/HCI

Plus Two Chemistry Alcohols, Phenols and Ethers Four Mark Questions and Answers

Question 1.
Match the following:

A	B
Oxidising Agent	Lucas Test.
Dehydrogenation	LiAIH4.
Anhydrous ZnCI2, HCI	Copper at 573K.
Reducing Agent	Acidified KMnO4.

Answer:

A	B
Oxidising Agent	Acidified KMnO4.
Dehydrogenation	Copper at 573K.
Anhydrous ZnCI2, HCI	Lucas Test.
Reducing Agent	LiAIH4.

Question 2.

a) Alcohols are having high boiling points than corresponding alkyl halides and ethers. Why?
b) Phenol is more acidic than ethanol. Give the reason.
c) Predict the products :

Answer:
a) Due to the presence of polar hydroxyl group alcohols can associate through intermolecular hydrogen bonding.
b) Phenoxide ion is stabilized by resonance while alkoxide ion has no resonance stabilisation.
c)

Question 3

1. Phenol is acidic even though it has no carboxylic group, why?
2. Convert phenol to salicylic acid and name the reaction.

Answer:
1. By the removal of a proton from phenol a phenoxide ion is obtained. It is stabilised by resonance. Hence phenol acts as an acid.
2.

.

Question 4.
1. Compare the solubility of diethyl ether and n- butane in water.
2. Give the product(s) from the reaction of one mole of diethyl ether with

• one mole of conc. HI and
• excess of HI.

Answer:
1. Diethyl ether being weakly polar is capable of forming intermolecular hydrogen bonding with water. Hence, diethyl ether is soluble in water while n-butane is not.

2. The product(s) from the reaction of one mole of diethyl ether.

• C₂H₅OH and C₂H₅I
• 2C₂H₅I

Question 5.
Identify X and Y.

Answer:

Question 6.
1. Boiling point depends on the inter molecular hydrogen bonding.

• Ethanol and propane have comparable molecular masses but their boiling points differ widely. Why?

2. Williamson synthesis is an important method of ether synthesis. Which of the following reactions is better for ether synthesis? Justify.

Answer:
1. Ethanol molecules are associated by intermolecular hydrogen bonding.

• Ethanol molecules are associated by intermolecular hydrogen bonding. This is absent in propane. So Ethanol has higher boiling point.

2. (CH₃)₃C-ONa + CH₃-Br is better. The tertiary alkyl halide undergoes elimination in presence of base to form alkene.

Question 7.
Chloro methane reacts with aqueous sodium hydroxide to form methanol.

1. What happens when chlorobenzene reacts with aqueous sodium hydroxide? Justify.
2. Write the reaction by which chlorobenzene can be converted to phenol.

Answer:

1. No reaction. This is due to sp² state of carbon to which CI is attached, less polarity of C-X bond and resonance stabilisation.
2. Dow’s process.
   When chlorobenzene is heated with sodium hydroxide solution at 623 K under a pressure of 200 atm in the presence of copper catalyst, sodium phenoxide is obtained. This on hydrolysis gives phenol.

Question 8.
Phenol exhibit acidic character.

1. Why phenol exhibit acidic character?
2. Explain it with the help of resonating structure of phenoxide ion.

Answer:

1. Phenol can donate proton and phenoxide ion thus formed is stabilized by resonance.
2. The resonating structure of phenoxide ion is given below.

Question 9.
What is the action of phenol with

1. Aqueous Br₂?
2. Br₂ in CS₂?
3. Nitrating mixture?
4. dil. HNO₃?

Answer:
1. Phenol on action with aqueous Br₂ gives 2, 4, 6,- tribromophenol.

2. Phenol on action with Br₂ in CS₂ gives o- bromophenol and p-bromophenol.

3.

4.

Question 10.
Write notes on
a) Kolbe’s reaction
b) Reimer-Tiemann reaction
Answer:
a) Kolbe’s reaction: When Sodium phenoxide is heated with CO₂ at 400 K and under a pressure 6-7 atm, sodium salicylate is obtained. This on hydrolysis gives ortho-hydroxy benzoic acid or salicylic acid.

Reimer – Tiemman reaction: When phenol is heated with CHCI₃ at 340 K, o-hydroxy benzaldehyde or salicylaldehyde is obtained.

Question 11.

1. Fill in the blanks:
   CH₃CH₂OH + SOCI₂ → CH₃CH₂CI+ …?… +….?….
2. This method is used to prepare extra pure alkyl halide. Do you agree? Why?

Answer:

1. CH₃CH₂OH + SOCI₂ -> CH₃CH₂CI + SO₂ + HCI
2. Yes. The by products are escapable gases. Hence, the reaction gives pure alkyl halides.

Question 12.
1. Arrange the compounds in the increasing order of their strength.

• 4-Nitro phenol
• Phenol
• Propan-1-ol
• 4-Methyl phenol.

2. You are given benzene, conc.H₂SO_4, and NaOH. Prepare phenol using these compounds.
Answer:
1. 4-Nitrophenol > Phenol > 4-Methylphenol > 1- Propanol. Presence of electron with darwing groups at ortho and para positions increases the acidic strength of substituted phenols.

2. By the action of benzene with conc.H₂SO₄ & NaOH, sodium benzene sulphonate is formed. This on acidification gives phenol.

Plus Two Chemistry Alcohols, Phenols and Ethers NCERT Questions and Answers

Question 1.
Explain why propanol has higher boiling point than that of the hydrocarbon, butane?
Answer:
Propanol has higher boiling point than butane because it has stronger interparticle forces. In propanol intermolecular hydrogen bonding is present whereas in butane intermolecular forces are weak van der Waals’ forces. A lot of heat is required to break intermolecular hydrogen bonding among propanol molecules.

Question 2.
Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
Answer:
Because of the presence of O-H group in them, alcohols are capable of forming hydrogen bonds with water molecules whereas hydrocarbons cannot form hydrogen bonds with water. As a result, alcohols are more soluble in water than the hydrocarbons of comparable molecular masses.

Question 3.
Give the structures and IUPAC names of monohydric phenols of molecular formula, C₇H₈O.
Answer:

Question 4.
Explain why ortho-nitrophenol is more acidic than ortho-methoxy phenol?
Answer:
ortho-Nitrophenol is more acidic than ortho methoxy phenol because nitro group by its electron with drawing resonance effect stabilises the phenoxide ion whereas methoxy group by its electron releasing effect destabilises the phenoxide ion. Greater the stability of the phenoxide ion, greater is the dissociation of phenol and greater is its acid strength.

Question 5.
Give IUPAC names of the following ethers:
i) C₂H₅OCH₂-CH(CH₃)CH₃
ii) CH₃OCH₂CH₂CI
iii) O₂N-C₆H₄-OCH₃(p)
iv) CH₃CH₂CH₂OCH₃

Answer:
i) 1-Ethoxy-2-methylpropane
ii) 2-Chloro-1-methoxyethane
iii) 4-Nitroanisole
iv) 1-Methoxypropane
v) 1-Ethoxy-4, 4-dimethyl cyclohexane
vi) Ethoxybenzene

Students can Download Chapter 6 Electro Magnetic Induction Questions and Answers, Plus Two Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Physics Chapter Wise Questions and Answers Chapter 6 Electro Magnetic Induction

Plus Two Physics Electro Magnetic Induction NCERT Text Book Questions and Answers

Question 1.
A horizontal straight wire 10m long extending from east to west is falling with a speed of 5.0ms^-1 at right angles to the horizontal component of the earth’s magnetic field 0.3 × 10^-4Wbm².

1. What is the instantaneous value of the e.m.f? induced in the wire?
2. What is the direction of the e.m.f.?
3. Which end of the wire is at the higher electrical potential?

Answer:
I – 10m, v = 5.0ms^-1, B = 0.30 × 10^-4Wbm²
1. Instantaneous value of e.m.f. induced
e = B lv
= 0.30 × 10^-4 × 10 × 5.0
or e = 1.5 × 10^-3V.

2. The direction of e.m.f. is from eastto west.

3. Since the current is flowing from east to west so the eastern end is at higher potential.

Question 2.
Current in a circuit falls from 5.0Ato 0.0A in 0.1s. If an average e.m.f. of 200V induced, give an estimate of the self-inductance of the circuit.
Answer:
dl = l₂ – l₁ = 0.0 – 5.0 = -5.0A,
dt = 0.1s, e = 200V, L=?
Since

or L = 4H

Question 3.
A pair of adjacent coils has a mutual inductance of 1 5H. If the current in one coil changes from 0 to 20A in 0.5s. What is the charge of flux linkage with the other coil?
Answer:
M= 1.5H, dl = l₂ – l₁ = 20 – 0 = 20A, dt = 0.5s, dΦ = ?
Since

And also

or dΦ = Mdl
or dΦ = 1.5 × 20 = 30 Wb.

Question 4.
A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing 25m long if the earth’s magnetic field at the location has a magnitude of 5.0 × 10^-4 and the dip angle is 30?
Answer:
v = 1800 km/h = 500ms^-1 I = 25m
Vertical component of B along horizontal
B = -Bsinθ = 5.0 × 10^-4 sin30° = 2.5 × 10^-4T
e = B/v sinθ
= 500 × 2.5 × 10^-4 × 25
= 31 V
The direction ofthe wing is immaterial.

Question 5.
Suppose the loop in Q.6.4 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.3T at the rate of 0.02TS^-1. If the cut is joined and the loop has a resistance of 1.6Ω, how much power is dissipated by the loops as heat? What is the source of this power?
Answer:
Induced e.m.f.

= 8 × 2 × 10^-4 × 0.02 = 3.2 × 10^-5V
Induced current,

= 2 × 10^-5A
Power Loss = I²r = (2 × 10^-5)² × 1.6 = 6.4 × 10^-10 W. Source of this power is the external agent responsible for changing the magnetic field with time.

Plus Two Physics Electro Magnetic Induction One Mark Questions and Answers

Question 1.
Eddy current are produced when.
(a) a metal is kept in varying magnetic field
(b) a metal is kept in steady magnetic field
(c) a circular coil is placed in a magnetic field
(d) through a circular coil, current is passed
Answer:
(a) a metal is kept in varying magnetic field

Question 2.
A100 milli henry coil carries a current of 1 A. energy stored in its magnetic field is
(a) 0.5 J
(b) 1 A
(c) 0.05 J
(d) 0.1 J
Answer:
(c) 0.05 J
Explanation : E = \(\frac{1}{2}\) LI² = \(\frac{1}{2}\) × (100 × 10^-3) × 1²
= 0.05 J.

Question 3.
A straight line conductor of length 0.4 m is moved with a speed of 7m/s perpendicular to a magnetic field of intensity 0.9 Wb/m². The induced e. m. f. across the conductor is.
(a) 5.04 V
(b) 25.2 V
(c) 1.26 V
(d) 2.52
Answer:
(d) 2.52
Induced e.m.f (s) = B/V
= 0.9 × 0.4 × 7 = 2.52V.

Question 4.
The core of a transformer is laminated because
(a) ratio of voltage in primary and secondary may be increased
(b) energy losses due to eddy currents may be minimized
(c) the weight of the transformer may be reduced
(d) rusting of the core may be prevented
Answer:
(b)

Plus Two Physics Electro Magnetic Induction Four Mark Questions and Answers

Question 1.

A conductor XY is moving through a uniform magnetic field of intensity B as shown in figure.

1. Name the emf.
2. The motion of XY towards right side causes an anticlockwise induced current. What will be the direction of magnetic induction in the region A?
3. The length of the conductor XY is 20cm. It is moving with a velocity 50m/s perpendicular to the magnetic fie Id. If the induced emf in the conductor is 2V find the magnitudes of magnetic field?

Answer:
1. Motional emf/induced emf.

2. Applying right hand grip rule at A direction of magnetic field is away from the paper.

3. ε = Blvsinθ, I = 20cm = 20 × 10^-2m, v
= 50m/s, ε = 2v

= 0.2 T.

Question 2.
1. When S₁ close bulb glows instantaneously when S₂ closes there is a delay in glowing the bulb.(1) This is due to………

• resistance of the coil
• back emf in the coil
• mutual induction
• none

2. Explain the phenomenon self-induction regarding above experiment (2)
3. Draw the graph with energy and current for a inductor. (1)
Answer:
1. Back emf in the coil.

2. When a.c. current is passed through the coil, a change in flux is produced. This change in flux produces a back e.m.f. in the coil. The phenomena of production of back emf is called self induction.

3. The graph with energy and current for a inductor:

Plus Two Physics Electro Magnetic Induction Five Mark Questions and Answers

Question 1.
When AC is switched on the thin metallic disc is found to thrown up in air.

1. Which makes the disc to thrown?
2. How will you explain the mechanism behind the movement of disc
3. Write the working principle of induction heater.

Answer:

1. Eddy current produced in the coil thrown disc into air.
2. Whenever the magnetic flux linked with a metal block changes, induced currents are produced due to this current, disc becomes a magnet. Hence disc thrown up in to air.
3. The change in flux produces eddy current in a metal. The heat produced by eddy current is used for cooking in induction heater.

Question 2.
AC generator is a device based on electromagnetic induction used to convert mechanical energy into electrical energy.

1. Draw a graph showing variation of induced emf over a cycle. Also indicate peak value of emf. (2)
2. How peak value of emf is related to its rms value? (1)
3. A rectangular wire loop of side 4cm × 6cm has 50 turns uniformly from 0.1 Tesla to 0.3 Tesla in 6 × 10^-2 second. Calculate induced emf in the coil. (2)

Answer:
1.

2. Erms = \(\frac{E_{0}}{\sqrt{2}}\)

3.

and Φ = BAN
Φ₁ = 0.1 × (24 × 10^-4) × 50
= 120 × 10^-4wb
Φ₂ = 0.3 × (24 × 10^-4) × 50
= 360 × 10^-4wb

= 40 × 10^-2V.

Students can Download Chapter 3 Production and Costs Questions and Answers, Plus Two Economics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Economics Chapter Wise Questions and Answers Chapter 3 Production and Costs

Plus Two Economics Production and Costs One Mark Questions and Answers

Question 1.
Which among the following cost curves is not ‘U’ shaped?
Answer:
AFC curve is rectangular hyperbola.

Question 2.
Identify the shape of the following AFC curve?
Answer:

The shape of AFC curve is rectangular hyperbola.

Question 3.
When a firm increased the number of labour from 10 to 11, keeping the capital fixed the total product increased from 120 to 130. Which of the following statement is correct in this regard?
(a) The total product fell
(b) This is a long run production
(c) The average product is rising
(d) 10 is the marginal product of an increased unit of labour.
Answer:
(d) 10 is the marginal product of an increased unit of labour.

Question 4.
Slope of an isoquant is:
(a) marginal cost
(b) DMRS
(c) DMRTS
(d) None of the above
Answer:
(c) DMRTS

Question 5.
When MP becomes zero, TP:
(a) increases
(b) decreases
(c) becomes maximum
(d) becomes negative
Answer:
(c) becomes maximum

Question 6.
Which among the following is not ‘u’ shape
(a) MC
(b) AC
(c) AVC
(d) AFC
Answer:
(d) AFC

Question 7.
When AC is minimum
(a) MC > AC
(b) MC = AC
(c) MC < AC
(d) None of these
Answer:
(b) MC = AC

Question 8.
All the following curves are U shaped except
(a) the AVC curve
(b) the AFC curve
(c) the AC curve
(d) the MC curve
Answer:
(b) the AFC curve

Plus Two Economics Production and Costs Two Mark Questions and Answers

Question 1.
Let the production function of a firm be Q = 5 L^1/2 K^1/2
Find the maximum output that the firm can produce with 100 units of L and 100 units of K.
Answer:
The production function is
Q = 5 L^1/2 K^1/2
Since the firm uses 100 units of L and 100 units of
K, we get the production function as
Q = 5 × 100^1/2 100^1/2
Q = 5 × 10x 10
Q = 5 × 100 = 500

Question 2.
Derive marginal product and average product from the total product schedule.

Answer:

Question 3.
Find the maximum possible output for a firm with zero units of labour and 10 units of capital when its production function is
Q = 5L + 2K
Answer:
The production function is Giving the values of L and K, we get
Q = 5 × 0 + 2 × 10
Q = 0 + 20 = 20

Question 4.
“In the long run, all costs are variable”. Do you agree? Justify your answer.
Answer:
Yes. In the long run, all costs are variable. This is because there is sufficient time available in the long run for any factor of production to get increased. As there are only variable factors, in the long run, all costs are variable.

Question 5.
“The AVC is U shaped” explain the reasons for the ‘U’ shape of AVC curve. Also, draw a diagram to clarify your points.
Answer:
The SMC and AVC curves start rising when production starts. As output increases, SMC falls. AVC being the average of marginal costs also falls but falls less than SMC. Then after a point, SMC start rising. AVC, however, continues to fall as long as value of SMC remains less than the prevailing value of AVC.

Once the SMC has risen sufficiently its value becomes greater than the value of AVC. The AVC starts rising. Therefore, the AVC curve is “U” shaped. It is given in the following diagram.

Question 6.
TFC of a firm is ₹2,000 and TVC is ₹3,000. It produces 20 units. Calculate the AVC and AC of the firm.
Answer:
TC = TFC + TVC
AC = 5000/20 = 250
AVC = 3000/20 = 150

Question 7.
Can there be some fixed cost in the long run?
Answer:
No, there cannot be some fixed cost in the long run. This is because, in the long run all factors of production can be adjusted and variable.

Question 8.
Let the production function of a firm be Q = 10 L^1/2 K^1/2 Find out the maximum possible output that the firm can produce with 100 units of L and 100 units of K.
Answer:
The production function is,
Q = 10 L^1/2 K^1/2
L =100 units
K =100 units
∴ Q = 10 × 1001/2 × 1001/2
= 10 × (10²)^1/2 × (10²)^1/2
= 10 × 10 × 10
= 1,000 units

Question 9.
The table given below shows the total product schedule of labour. Determine the AP and MP of labour?

Answer:

Question 10.
Prove that AC is the sum of AFC and AVC.
Answer:

Question 11.
A bus company started its operation with two buses, four labourers. It has paid a one-time road tax. As the passengers increased the company arranged more tripes, with the help of additional labour and for using more fuels. Classify the costs incurred by the bus company into fixed and variable.
Answer:

Question 12.
Both short run and long run average and marginal cost curves are U-shaped. But reasons for the U-shape are not the same. Bring out the differences.
Answer:
Behaviour of output in the short run as explained by the law of variable proportions Behaviour of output in the long run as explained by the laws of returns to scale.

Plus Two Economics Production and Costs Three Mark Questions and Answers

Question 1.
Match the following.

Answer:

Question 2.
Categorize the following into fixed cost
(a) wages of temporary workers
(b) cost of raw materials
(c) salary of permanent staff
(d) cost of transportation
(e) cost of plant
Answer:

Fixed Cost	Variable Cost
Salary of permanent staff	Cost of raw materials
Cost.of acquiring land	Cost of transportation
Cost of plant	Wages of temporary workers

Question 3.
Identify the following curves.

Answer:

1. TC
2. TVC
3. TFC

Question 4.
Identify the shapes of the following cost curves.

1. AFC
2. AC
3. TFC
4. TVC
5. TC

Answer:

1. AFC – Rectangular hyperbola
2. AC – ‘U’shape
3. TFC – Straight line parallel to X axis
4. TVC – Inverse‘S’ shape
5. TC – Inverse‘S’ shape

Question 5.
There exists vertical distance between 1. TVC and TC and (b) TC and TFC. What does this distance indicate?

1. TVC and TC
2. TC and TFC

Answer:

1. The vertical distance between TVC and TC represents the Total Fixed Costs (TFC).
2. The vertical distance between TC and TFC represents the Total Variable Costs (TVC)

Question 6.
Complete the following table

Answer:

Question 7.
Name the following curves. ,

Answer:

Question 8.
What do you mean by an ‘isoquant’?
Answer:
An isoquant is the set of all possible combinations of the two inputs that yield the same maximum possible level of output. Each isoquant represents a particular level of output and is labelled with that amount of output. The shape of isoquant is drawn below.

Question 9.
With the help of a diagram show the relation between average exists and average variable cost.

Answer:
Both AC and AVC are U shaped. As the output increases the gap between AC and AVC narrows.

Question 10.
In the short run, Total Variable Cost is zero when output is zero. When output rises, Total Cost also rises. Draw a suitable diagram and explain the relationship between Total Fixed Cost, Total Variable Cost and Total Cost.
Answer:

Question 11.
“Average Fixed Cost (AFC) curve is a continuously falling curve.”

1. Substantiate this statement by giving the reasons.
2. Graphically represent the AFC curve

Answer:
1. TFC is constant & hence AFC falls
2.

Plus Two Economics Production and Costs Five Mark Questions and Answers

Question 1.
The marginal product and total product of an input are related’. Prove this statement.’
Answer:
The marginal product (MP) and total product (TP) of an input are related. The points of relationship are given below.

1. When MP increases, TP also increases
2. When MP is zero, TP becomes maximum
3. When MP becomes negative, TP turns negative

Question 2.
Differentiate between returns to a factor and returns to scale.
Answer:
Returns to a factor refers to the effects on output of changes in one input with all other inputs are hold constant. On the other hand, returns to scale refers to the effect on total output of charges in some constant rate in all the inputs simultaneously.

Question 3.
The marginal cost and average cost are related to each other. Prove this using diagram and a numerical example?
Answer:

Marginal cost are addition made to the total costs by the production of an additional unit of the commodity.
MC = TC_n – TC_n-1

Average lost is per unit cost of production which is obtained by dividing the total cost by number of units of output produced.
\(A C=\frac{T C}{Q}\)

Since AC is obtained by dividing TC by all units, MC is the addition to TC by producing an additional unit so MC brings a change in AC.

From above:

1. When MC is less than AC, then MC will fall
2. When MC is equal to AC, AC remains costant.
3. When MC is more than AC, AC rises

Question 4.
Which of the following represent the long run production function?
1. MC is less than AC, then MC will fall

• Law of variable proportion
• Returns to scale

2.  Explain what may happen when all inputs vary simultaneously?
Answer:
Returns to scale.
When all inputs are varied simultaneously the output may change in three ways.
1. Constant returns to scale:
It is when a proportional increase in inputs result in an increase in output by the same proportion.

2. Increasing returns to scale:
It is when a proportional increase in all inputs result in an increase in output by more than proportion.

3. Decreasing returns to scale:
It is when a proportional increase in all inputs result in an increase in output by less than proportion

Question 5.
State whether the statement are true or false.

1. TC never becomes zero.
2. AC is sum total of AFC and AVC.
3. When AC rises, AC and MC are equal
4. Real cost is the cost in money terms
5. TFC curve is U shaped

Answer:

1. True
2. True
3. False
4. False
5. False

Question 6.
The following table shows the total cost schedule of a firm. What is the total fixed cost schedule of the firm? Calculate the JVC, AFC, AVC, SAC and SMC of the firm

Answer:

Question 7.
The diagram shows the total product curve of a factor in the law of variable proportion.

1. Explain the law
2. With the help of a diagram illustrate the average and marginal product curve and their relation.

Answer:
1. Law of variable proportion says that if one variable input is added with other fixed inputs the marginal product of a factor input initially rises but after reaching a certain level of output it starts falling.

2. The relation between the average and marginal product is given below. BothAPand MPare ‘n ’shaped. MPwill always pass through the maximum of AP.

Question 8.
Theory of production deals with producer’s behaviour, and according to this theory output produced by a firm passes through three stages in the short run.

1. Which are the three stages of production?
2. Analyse and bring out the salient features of each stage.
3. Which stage of production is very important as far as a firm is concerned? Why?
4. Give suitable diagram by drawing the Total Prod-uct, Average Product & Marginal Product curves.

Answer:

1. Increasing, diminishing and negative
2. First output increases at increasing rate, then at diminishing rate & finally starts falling.
3. Second stage, profit maximisation occurs in this Stage.
4.

Question 9.
If other things remaining same, graphically explain what happens to the supply curve for readymade shirts if there is

1. An increase in the wages paid to the tailors.
2. An increase in the price of ready-made shirts

Answer:

Plus Two Economics Production and Costs Eight Mark Questions and Answers

Question 1.
Prepare a seminar report on the topic “Production Function.”
Answer:
Respected teachers and dear friends:
The topic of my seminar paper is “Production Function”. The production function of a firm shows relationship between inputs and output. In this seminar paper, I would like to present the different production functions such as short run production function and the long run production function.

Introduction:
As we know the production function shows the transformation of inputs into output. The production function can be of two types – short run production function known as Law of Variable proportion and the long run production function known as returns to scale. Q =F(F₁ F₂… F_n)

Contents:
A. Law of Variable proportion

1. Increasing returns
2. Diminishing returns
3. Negative returns

A. The Law of Variable Proportions
When more and more units of a variable input are added with the fixed input, the marginal product would increase only upto a certain point. Therefore, the marginal product declines. This phenomenon is known as the Law of Variable Proportions. It is also known as returns to a factor.

The shape of TP, AP and MP suggests that they are specifically passing through three phases.
They are:
1. First phase:
In the first stage, both AP and MP increase. As a result TP also increases at an increasing rate. This stage is known as the stage of increasing return to a factor. AP reaches the maximum level in this stage.

2. Second phase:
Both AP and MP decrease at this stage. The TP increases at a decreasing rate. More importantly, TP reaches maximum and MP touches zero. This stage is also known as the stage of diminishing returns to a factor.

3. Third phase:
At this stage, the MP becomes negative. As a result, TP also starts declining. The decline of AP is continuous. In the graph, when TP reaches maximum and MP touches zero. When MP becomes negative, TP starts declining. This stage is known as the stage of negative returns to a factor.

B. Returns to scale

1. Increasing returns
2. Constant returns
3. Decreasing returns
   (for details, refer summary part)

B. Returns to scale
As stated earlier, all the factor inputs become variable in the long run and thereby no distinction can be made between fixed inputs and variable inputs. Return to scale is associated with long run production function.

Returns to scale refers to the change in output when all inputs are variable and the proportion between inputs remains constant. When all the inputs vary in the same proportion, the output (TP) behaves in different manner, which can be clubbed into three categories.

1. Increasing Returns to Scale (IRS):
When a proportionate change in all the inputs leads to more than proportionate change in output, it is known as the stage of increasing returns to scale. For instance, a 10% change in inputs results in more than 10% change in output.

2. Constant Returns to Scale (CRS):
When a proportionate change in all the inputs leads to change in output in the same proportion is known as the stage of constant returns to scale. It indicates that a 10% change in inputs leads to exactly 10% change in output.

3. Decreasing Returns to Scale (DRS):
When a proportionate change in all the inputs leads to less than proportionate change in output is known as the stage of decreasing returns to scale. It indicates that a 10% change in inputs leads to less than 10% change in output.

Conclusion:
Thus it can be concluded that there are two types of production function depending upon the use of inputs and the time period.

Question 2.
Prepare a seminar paper on Law of Returns to Scale.
Answer:
Returns to scale:
As stated earlier, all the factor inputs become variable in the long run and thereby no distinction can be made between fixed inputs and variable inputs. Return to scale is associated with long run production function.

Returns to scale refers to the change in output when all inputs are variable and the proportion between inputs remains constant. When all the inputs vary in the same proportion, the output (TP) behaves in different manner, which can be clubbed into three categories.

1. Increasing Returns to Scale (IRS):
When a proportionate change in all the inputs leads to more than proportionate change in output, it is known as the stage of increasing returns to scale. For instance, a 10 percentage change in inputs results in more than 10 percentage change in output.

2. Constant Returns to Scale (CRS):
When a proportionate change in all the inputs leads to change in output in the same proportion is known as the stage of constant returns to scale. It indicates that a 10 percentage change in inputs leads to exactly 10 percentage change in output.

3. Decreasing Returns to Scale (DRS):
When a proportionate change in all the inputs leads to less than proportionate change in output is known as the stage of decreasing returns to scale. It indicates that a 10 percentage change in inputs leads to less than 10 percentage change in output.

Question 3.
From the table given below

1. Fill the blank columns
2. Using the information in the table draw TC, TVC, TFC in one and AC, AVC, AFC, MC in other and prepare a short note on them.

Answer:
1.

2. Use the data to draw the diagram.
TC, TVC and TFC represent total costs. TC = TFC + TVC. TFC is the cost incurred on fixed factors. TVC is the cost incurred on variable costs. Even though the level of output is zero there will be some costs. This is known as fixed cost. When the level of output is zero total variable costs also will be zero.

It will increase as the level of output increases. AFC is the cost that always falls. It is a rectangular hyperbola. AVC, AC, MC are ‘U’ shaped. The MC will always pass through the minimum oftheAVCand AC.

Students can Download Chapter 12 Aldehydes, Ketones and Carboxylic Acids Questions and Answers, Plus Two Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Chemistry Chapter Wise Questions and Answers Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Plus Two Chemistry Aldehydes, Ketones and Carboxylic Acids One Mark Questions and Answers

Question 1.
The reaction is called
(a) Sandmeyer’s reaction
(b) Rosenmund’s reduction
(c) HVZ reaction
(d) Cannizaro’s reaction
Answer:
(b) Rosenmund’s reduction

Question 2.
Say TRUE or FALSE:
Aldol condensation is given by all aldehydes and ketones.
Answer:
False

Question 3.
A 40% solution of _________ is called formation
Answer:
formaldehyde (HCHO)

Question 4.
Benzamide on heating with bromine and caustic alkali gives
(a) benzene
(b) methylamine and benzene
(c) aniline
(d) m-Bromobenzaldehyde
Answer:
(c) aniline

Question 5.
In the reaction the product ‘B’ is
(a) acetanilide
(b) glycine
(c) ammonium acetate
(d) methane
Answer:
(b) glycine

Question 6.
Arrange the following in the decreasing order of acidity.
ClCH₂COOH, Cl₃CCOOH, CH₃COOH, Cl₂HCOOH
Answer:
Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH > CH₃COOH

Question 7.
Reaction of butanone with methyl magnesium bromiode followed by hydrolysis gives_________
Answer:
2 methyl -2- butanol

Question 8.
The major product of the addition of water molecule to propyne in the presence of mercuric sulphate and dil sulphuric acid is ________
Answer:
Propanone

Question 9.
One mole of propanone and one mole of formalde¬hyde are the products of ozonolysis of one mole of an alkene. The alkene is ________
Answer:
2 methyl propene

Question 10.
Which of the following is a better reducing agent for the following reduction.
RCOOH → RCH₂OH
(a) SnCI₂/HCI
(b) NaBH₄/ether
(c) H₂/pd
(d) N₂H₄/C₂H₅ONa
(e) B₂H₆/H₃O⁺
Answer:
(e) B₂H₆/H₃O⁺

Question 11.
The total number of acyclic structural isomers possible for compound with molecular formula C₄H₁₀O is ________
Answer:
7

Plus Two Chemistry Aldehydes, Ketones and Carboxylic Acids Two Mark Questions and Answers

Question 1.
Reactivity of ketone is less than that of aldehyde. Why?
Answer:
Due to steric hindrance and inductive effect of alkyl group.

Question 2.
Carboxylic acid decompose into carboxylate ion and H⁺ ion.

1. Explain this on the basis of resonating structure of carboxylic acid.
2. Arrange the following in the increasing order of acidity. HCOOH, CH₃COOH
3. Substantiate.

Answer:

1. Carboxylic acid decomposes to give proton and carboxylate ion and is stabilized by resonance. This explains the acidic character of carboxylic acid.
2. CH₃COOH < HCOOH
3. In acetic acid the electron donating effect (+l – effect) of -CH₃ group destabilises the carboxylate anion and decreases the acid strength. Whereas in formic acid the H atom has not electron withdrawing or electron donating effect.

Question 3.
What is Etard’s reaction?
Answer:
Etard’s reaction:
When toluene is oxidized using chromyl chloride, benzaldehyde is obtained.

Question 4.
What is HVZ reaction? Explain.
Answer:
HVZ reaction – When a carboxylic acid is treated with red P and halogen, the α-H atoms are successively replaced by halogen.

This reaction has great synthetic importance as the halogen atom can be replaced by a number of other groups giving useful products.

Question 5.
Predict the product and name the reaction:

1. HCHO + NaOH → B + C
2. CH₃COOH + CH₃OH → E

Answer:

1. CH₃OH + HCOONa – Cannizzaro reaction
2. CH₃COOCH₃ – Esterification

Question 6.
Write the name of any two tests to distinguish between acetaldehyde and acetone.
Answer:
Benedict’s test, Fehling’s test – Both tests are answered by acetaldehyde and not by aceotne.

Question 7.
An organic compound with molecular formula C₉H₁₀O forms 2, 4-DNP derivative, reduces Tollens’reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1, 2-benzenedicarboxylic acid. Identify the compound.
Answer:
From the given data it is clear that as the compound forms2, 4-DNP derivative it has /CO group. Since it reduces Tollens’ reagent therefore -CHO group is present. As it can also undergo Cannizzaro reaction therefore α -hydrogen is absent

The oxidation product suggests that the compound has a benzene ring. One of the – COOH groups have been obtained by the oxidation of – CHO group and the other from alkyl group. Hence on these basis, the structure of C₉H₁₀O is

Question 8.
Write a notes on

1. Reimer – Tiemann reaction
2. Rosenmund Reduction

Answer:
1. Reimer – Tiemman reaction: When phenol is heated with CHCI₃ at 340 K, o-hydroxy benzaldehyde or salicylaldehyde is obtained.

2. Rosenmund reduction: When an acid chloride is reduced by using hydrogen gas in presence of Pd and BaSO₄, an aldehyde is obtained.
OR

Question 9.
Distinguish the following compounds using any one test.
H₃C – CO – CH₃ and CH₃CH₂CHO
Answer:
CH₃COCH₃ give Iodoform test which CH₃CHO does not answer this test.

Question 10.
Aspirin is commonly used in medicine. How it is prepared? Give the equation.
Answer:
Aspirin is acetyl salicyclic acid. When salicyclic acid is treated with acetyl chloride, aspirin is obtained.

Salicylic acid + acetyl chloride → Aspirin

Question 11.
How will you prepare CH₃-CO-NH₂ and CH₃COOCH₃ from CH3COOH?
Answer:

Question 12.
Give the IUPAC name of the following compounds.
i) C₆H₅CH = CHCHO
ii) (CH₃)₃CCH₂COOH
Answer:

Question 13.
Give a test to distinguish between acetaldehyde and acetone.
Answer:
CH₃ – CO – CH₃ contains CH₃CO – group and hence it gives iodoform test.
CH₃ – CH₂ – CHO does not give iodoform test.

Question 14.
Predict the major product in the following reactions:

Answer:

Question 15.
Distinguish between formaldehyde & acetaldehyde.
Answer:

HCHO	CH3CHO
1. It gives Cannizzaro reaction.	1. It doesn’t give Cannizzaro reaction.
2. It doesn’t give aldol condensation.	2. It gives aldol condensation.
3. It gives condensation products with NH3.	3. It gives addition products with NH3.

Question 16.
Which is more acidic, 2-chloropropanoic acid or 3- chloropropanoicacid? Why?
Answer:
2-chloropropanoic acid. Becasue the electron with drawing -Cl group is more closer to -COOH group in this compound.

Plus Two Chemistry Aldehydes, Ketones and Carboxylic Acids Three Mark Questions and Answers

Question 1.
Fill in the blanks:

Answer:

Question 2.
A student is given Tollens’ reagent for oxidation of aldehyde.

1. What is Tollens’ reagent?
2. Can you help him to do the experiment?
3. What is the result of the experiment?

Answer:

1. Tollens’ reagent is ammoniacal silver nitrate solution
2. Yes. To a little of the solution add Tollens’ reagent
3. A black precipitate of silver or silver mirror is obtained

Question 3.

1. What is the role of LiAIH₄?
2. Give one example of an oxidising agent?
3. What is the action of a carboxylic acid with alcohol?

Answer:

1. LiAIH₄ is a strong reducing agent. It reduces RCOOH to 1° alcohol (RCH₂OH)
2. Acidified KMnO₄.
3. When carboxylic acid is treated with an alcohol in the presence of dehydrating agent like conc.H₂SO₄, an ester is formed. This is called esterification reaction.

Question 4.
What happens when primary, secondary and tertiary alcohols are passed over red hot copper? Give equations.
Answer:
1° alcohol hot copper Aldehyde
2° alcohol over hot copper – Ketone
3° alcohol over hot copper- alkene

Question 5
Fill in the blanks.

Answer:

1. HVZ reaction
2. CH₃CH₃
3. Hoffmann bromamide degradation reaction
4. HCOONa + CH₃OH
5. Reimer-Tiemann reaction
6. C₆H₅Cl

Question 6.
Draw the structure of the following derivatives.
(i) The 2,4-dinitrophenylhydrazone of benzaldehyde
(ii) Cyclopropanone oxime
(iii) Acetaldehydedimethylacetal
(iv) The semicarbazone of cyclobutanone
(v) The ethylene ketal of hexan-3-one
(vi) The methyl hemiacetal of formaldehyde
Answer:

Question 7.

i) 2HCHO + NaOH → CH₃OH + HCOONa
ii)

(a) Identify Cannizzaro and Aldol condensation.
(b) What is the difference between the above two reactions?

Answer:
(a) Cannizzaro reaction:
2HCHO + NaOH CH₃OH + HCOONa
Aldol condensation:

(b) Cannizzaro reaction is given by aldehydes having no α-H atom whereas aldol condensation is given by aldehydes containing α-H atom.

Question 8.
a) In a practical class a group of students heated ethanal with NaOH. Another group heated methanal with conc.NaOH.
i) Identify the products in each reaction with equation.
ii) Name the reactions.
b) Aldehydes are more reactive than ketones. Comment on the statement.
Answer:
a)

ii) The first reaction is Cannizarro reaction and the second reaction is aldol condensation.

b) Due to the ‘+l’ effect and steric hindrance of surrounding alkyl group ketones are less reactive than aldehydes.

Plus Two Chemistry Aldehydes, Ketones and Carboxylic Acids Four Mark Questions and Answers

Question 1.
i) Arrange the following in the increasing order of acidic strength and justify your answer.
CH₃COOH, CHCI₂COOH, CH₂CICOOH, CCI₃COOH

ii) Suggest a method to convert acetic acid to chloroacetic acid. Name the reaction and write the chemical equation.
Answer:
i) CH₃COOH < CH₂CI-COOH < CHCI₂COOH < CCI₃COOH This is due to the electron withdrawing character of chlorine.

ii) HVZ reaction – When a carboxylic acid is treated with red P and halogen, the a-H atoms are successively replaced by halogen.

This reaction has great synthetic importance as the halogen atom can be replaced by a number of other groups giving useful products.

Question 2.
a) Which of the following carbonyl compounds answer aldol condensation reaction and give equation.
HCHO, CH₃CHO, CCI₃CHO, C₆H₆-CHO

b) Arrange the following compounds in the increasing order of acidity:
CH₃COOH, CH₂CICOOH, CH₃-CH₂-COOH, C₆H₅-COOH
Answer:
a) Among these compounds only CH₃CHO answer aldol condensation reaction. Others will not answer this reaction because they have no α – hydrogen atom.

Question 3.

1. Aldehydes and ketones are carbonyl compounds Give a test for identification of aldehydes.
2. Acidic strength is related to the stability of carboxylate anion. Which acid of each pair is stronger?

Answer:
1. Benedict test. Benedict reagent is a mixture of sodium carbonate and sodium citrate. This on reaction with aldehyde gives red precipitate of Cu₂S.

2. Acidic strength is related to the stability of carboxylate anion. Acid of each pair is stronger:
i) CH₂FCOOH. This is due to the high electron with drawing effect of F.

This is due to the high electron with drawing effect of the -CF₃ group.

Question 4.
Substituents on carboxylic acids have much effect on their acidity. Substantiate the statement with the following examples.
a) HCOOH, CH₃COOH, CH₂CICOOH

Answer:
CH₂CICOOH > HCOOH > CH₃COOH
The methyl group will intensify the negative charge on the carboxylate ion and destabilise it as compared to formate ion. So HCOOH is stronger than CH₃COOH. The electron withdrawing effect of a Cl makes chloroacetic acid stronger than HCOOH and CH₃COOH.

In the case of aromatic carboxylic acids, presence of electron withdrawing groups at ortho and para position increases their acidity while presence of electron donating groups decreases their acidity.

In 4-nitro benzonic acid acid strength is greater than that of benzoic acid due to the electron withdrawing nature of -NO₂ group while in 4-methoxy benzoic acid acid strength decreases due to the electron donating nature of the methoxy group.

Question 5.
Give a chemical test to distinguish between each of
the following pair of organic compounds.

1. Propanal and Propanol
2. Propanone and Propanal

Answer:

1. Propanal is an aldehyde and it gives a silver mirror with Tollens’ reagent while propanol is an alcohol and will not answer Tollens’test.

2. Propanone gives yellow precipitate of iodoform on reaction with I₂ and NaOH while propanal does not give iodoform test. OR Propanal gives silver mirror with Tollens’ reagent while propanone does not give silver mirror test.

Question 6.
What is meant by the following terms? Give an example of the reaction in each case.

1. Cyanohydrin
2. Acetal
3. Semicarbazone
4. Aldol
5. Hemiacetal
6. Oxime
7. Ketal
8. Imine
9. 2,4-DNP-derivative
10. Schiff’s base

Answer:
1. Cyanohydrin – It is a compound which contain both OH and CN groups. For example, Lactic acid can be obtained by hydrolysis of cyanohydrin.

2. Acetal – compounds formed by the reaction of aldehydes with monohydric alcohols in presence of dry HCI gas.

3. Semicarbazone – the product of carbonyl compounds with semicarbazide is known as semicarbazone.

4. Aldol – It is a condensation product of aldehydes or ketones having atleast one α – hydrogen atom with dilute alkali as catalyst.

5. Hemiacetal – It is a compound which contains an ether as well as alcohol functional group. For example, methoxyethanol is a hemiacetal.

6. Oxime – Addition compound formed by the reaction of aldehyde or ketone with hydroxylamine.

7. Ketal – It is a cyclic compound obtained by reaction of aceone with ethylene glycol.

8. Imine – Addition compound formed by the reaction of aldehyde or ketone with ammonia.

9. 2, 4-DNP derivative – 2, 4-phenylhydrazone (DNP) is the addition compound formed by the reaction of aldehydes and ketones with 2, 4-dinitrophenylhydrazine.

10. Schiff’s base – Addition compound formed by the reaction of aldehyde or ketone with amine.

Question 7.
Name the following compounds according to IUPAC system of nomenclature.

1. CH₃CH(CH₃)CH₂CH₂CHO
2. CH₃CH₂COCH(C₂H₅)CH₂CH₂CI
3. CH₃CH=CHCHO
4. CH₃COCH₂COCH₃
5. CH₃CH(CH₃)CH₂C(CH₃)₂COCH₃
6. (CH₃)₂CCH₂COOH
7. OHCC₆H₅CHO-p

Answer:

1. 4-Methylpentanal
2. 6-Chloro-4-ethylhexan-3-one
3. But-2-enal
4. Pentane-2,4-dione
5. 3, 3, 5-Trimethylhexane-3-one
6. 3, 3-Dimethylbutanoicacid
7. Benzene 1, 4-dicarbaldehyde

Question 8.

1. What is the relation between an electron donating group and acidic character?
2. How carboxylic acids maintain their acid character?

Answer:

1. Electron donating group decreases the acid character.
2. Carboxylic acid decomposes to give proton and carboxylate ion and is stabilized by resonance. This explains the acidic character of carboxylic acid.

Question 9.
Predict the product formed when cyclohexane carbaldehyde reacts with following reagents:
i) PhMgBr and then H₃O⁺
ii) Tollens’reagent
iii) Semicarbazide and weak acid
iv) Excess ethanol and acid
v) Zinc amalgam and dilute hydrochloric acid
Answer:

Plus Two Chemistry Aldehydes, Ketones and Carboxylic Acids NCERT Questions and Answers

Question 1.
What is meant by the following terms? Give an example of the reaction in each case

1. Cyanohydrin
2. Acetal
3. Semicarbazone
4. Aldol
5. Hemiacetal
6. Oxime
7. Ketal
8. Imine
9. 2,4-DNP-derivative
10. Schiff’s base

Answer:
1. Cyanohydrin – It is a compound which contain both OH and CN groups. For example, Lactic acid can be obtained by hydrolysis of cyanohydrin.

2. Acetal – compounds formed by the reaction of aldehydes with monohydric alcohols in presence of dry HCI gas.

3. Semicarbazone – the product of carbonyl compounds with semicarbazide is known as semicarbazone.

4. Aldol – It is a condensation product of aldehydes or ketones having atleast one α – hydrogen atom with dilute alkali as catalyst.

5. Hemiacetal – It is a compound which contains an ether as well as alcohol functional group. For example, methoxyethanol is a hemiacetal.

6. Oxime – Addition compound formed by the reaction of aldehyde or ketone with hydroxylamine.

7. Ketal – It is a cyclic compound obtained by reaction of aceone with ethylene glycol.

8. Imine – Addition compound formed by the reaction of aldehyde or ketone with ammonia.

9. 2, 4-DNP derivative – 2, 4-phenylhydrazone (DNP) is the addition compound formed by the reaction of aldehydes and ketones with 2, 4-dinitrophenylhydrazine.

10. Schiff’s base – Addition compound formed by the reaction of aldehyde or ketone with amine.

Question 2.
Name the following compounds according to IUPAC system of nomenclature.

1. CH₃CH(CH₃)CH₂CH₂CHO
2. CH₃CH₂COCH(C₂H₅)CH₂CH₂CI
3. CH₃CH = CHCHO
4. CH₃COCH₂COCH₃
5. CH₃CH(CH₃)CH₂C(CH₃)₂COCH₃
6. (CH₃)₃CCH₂COOH
7. OHCC₆H₅CHO-p

Answer:

1. 4-Methylpentanal
2. 6-Chloro-4-ethylhexan-3-one
3. But-2-enal
4. Pentane-2,4-dione
5. 3, 3, 5-Trimethylhexane-3-one
6. 3, 3-Dimethylbutanoicacid
7. Benzene 1, 4-dicarbaldehyde

Question 3.
Draw the structure of the following compounds
(i) 3-Methylbutanal
(ii) p-Nitropropiophenone
(iii) p-methylbenzaldehyde
(iv) 4-methylpent-3-en-2-one
(v) 3-bromo-4-phenylpentanoicacid
(vi) 4-Chloropentan-2-one
(vii) p, p-Dihydroxybenzophenone
(viii) Hex-2-en-4-ynoicacid
Answer:

Question 4.
An organic compound with molecular formula C₉H₁₀O forms 2, 4-DNP derivative, reduces Tollens’reagent and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1, 2-benzene dicarboxylic acid. Identify the compound.
Answer:
From the given data it is clear that as the compound forms 2, 4-DNP derivative it has >CO group. Since it reduces Tollens’ reagent -CHO group is present. As it can also undergo Cannizzaro reaction α- hydrogen is absent.

The oxidation product suggests that the compound has a benzene ring. One of the – COOH groups have been obtained by the oxidation of – CHO group and the other from alkyl group. Hence on these basis, the structure of C₉H₁₀O is

Question 5.
Write structural formulas and names of the four possible aldol condensation products from propanal and butanal. In each case, indicate which aldehyde served as nucleophile and which as electrophile.
Answer:
i) Propanal as electrophile as well as nucleophile

ii) Propanal as electrophile and butanal as nucleophile

iii) Butanal as electrophile and propanal as nucleophile

iv) Butanal as electrophile as well as nucleophile

Question 6.
Predict the product formed when cyclohexane carbaldehyde reacts with following reagents:
i) PhMgBr and then H₃O⁺
ii) Tollens’reagent
iii) Semicarbazide and weak acid
iv) Excess ethanol and acid
v) Zinc amalgam and dilute hydrochloric acid
Answer:

Question 7.
Give simple chemical tests to distinguish between

1. Propanal and propanone
2. Acetophenone and Benzophenone
3. Phenol and Benzoic acid
4. Benzaldehyde and acetophenone
5. Ethanal and propanal

Answer:
1. Propanal and propanone:
Propanal and propanone can be distinguished by iodoform test as it is given by propanone and not by propanal
Propanone reacts with hot NaOH/I₂ to give yellow precipitate of iodoform.

2. Acetophenone and benzophenone:
Acetophenone gives iodoform test but benzophenone does not respond.

3. Phenol and benzoic acid:
This can be distinguished by treating FeCI₃ solution. Phenol gives violet colour with FeCI₃ solution while benzoic acid gives buff colured precipitate.

4. Benzaldehyde and acetophenone:
Acetophenone responds to iodoform test while benzaldehyde does not.

5. Ethanal and propanal:
Ethanal gives yellow ppt. of iodoform with an alkaline solution of iodine (iodoform test)

Propanal does not give yellow ppt.

Students can Download Chapter 5 Magnetism and Matter Questions and Answers, Plus Two Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Physics Chapter Wise Questions and Answers Chapter 5 Magnetism and Matter

Plus Two Physics Magnetism and Matter NCERT Text Book Questions and Answers

Question 1.
Answer the following questions regarding earth’s magnetism:

1. A vector needs three quantities for its specification. Name the three independent quantities conventionally used to specify the earth’s magnetic field.
2. The angle of dip at a location in southern India is about 18°. Would you expect a greater or lesser dip angle in Britain?
3. If you make a map of magnetic field lines at Melbourne in Australia, would the lines seen to go into the ground or come out of the ground?
4. In which direction would a compass free to move in the vertical plane point to, if located right on the geomagnetic north or south pole?
5. The earth’s field is claimed roughly approximately, the field due to a dipole magnetic moment 8 × 10²² JT^-1 located at the centre. Check your or¬der of magnitude of this number same way.
6. Geologists claim that besides the main magnetic NS poles, there are several local poles on the earth’s surface oriented in different directions. How is such a thing possible at all?

Answer:
1. Magnetic declination, angle of dip, horizontal component of earth’s magnetic field.

2. Greater angle of dip in Britain, (it is about 70°), because Britain is closer to the magnetic north pole.

3. Field lines of B due to the earth’s magnetism would seem to come out of the ground.

4. A compass is free to move in a horizontal plane, while the earth’s field is exactly vertical at the magnetic poles. So the compass can point in any direction there.

5. Use the formula for field B on the normal bisector of a dipole of magnetic moment m.
B = \(\frac{\mu_{0}}{4 \pi} \frac{M}{r^{3}}\)
Take M = 8 × 10²² JT^-1
r = 6.4 × 10⁶ m;
one gets B = 0.3G
which checks with the orde of magnitude of the observed field on the earth.

6. The earth’s field is only approximately a dipole field. Local N-S poles may arise due to, for instance, magnetized minerals deposits.

Question 2.
A short bar magnet of magnetic moment m = 0.32JT^-1 is placed in a uniform external magnetic field of 0.15T. If the bar is free to rotate in the plane of the field, which orientations would correspond to its

• stable and
• unstable equilibrium? What is the potential energy of the magnet in each case?

Answer:
Given M=0.32JT^-1 B = 0.15T, U = ?

• If \(\overrightarrow{\mathrm{M}} \| \overrightarrow{\mathrm{B}}\), then we have stable equilibrium and U = -MB = -0.32 × 0.15T = -4.8 × 10^-2 J
• If \(\overrightarrow{\mathrm{M}}\) is anti-parallel to \(\overrightarrow{\mathrm{B}}\) , we have unstable equilibrium and
  U = MB = 0.32 × 0.15J = 4.8 × 10^-2J

Question 3.
A closely wound solenoid of 800 turns and area of cross-section 2.5 × 10^-4 m² carries a current of 3.0A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?
Answer:
Given N = 800, A = 2.5 × 10^-4 m², I = 3.0A, M = ?
Since M = NIA
∴ M = 800 × 3 × 2.5 × 10^-4
or M = 0.60JT^-1 along the axis of the solenoid.

Question 4.
A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10^-4 m², carrying a current of 4.0A, is suspended through its centre allowing it to turn in a horizontal plane.

1. What is the magnetic moment associated with the solenoid?
2. What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 × 10^-2 is set up at an angle of 30° with the axis of the sole¬noid?

Answer:
1. Magnetic dipole moment,
M = nIA
= 2000 × 4.0 × 1.6 × 10^-4 = 1.28JT^-1

2. Net force = 0
Torque, τ = MB sinθ
= 1.28 × 7.5 × 10^-2 × sin 30°
= 1.28 × 7.5 × 10^-3 × 1/2 = 4.8 × 10^-2 Nm.

Question 5.
Answer the following questions:

1. Why does a paramagnetic sample display greater magnetization (for the same magnetizing field) when cooled?
2. Why is diamagnetism, in contrast, almost independent of temperature?
3. If a toroid used bismuth for its core, will the field in the core be (slightly) greater or (slightly) less than when the core is empty?
4.  If the permeability of a ferromagnetic material independent of the magnetic field? If not is it more for lower or higher fields?
5. Magnetic field lines are always nearly normal to the surface of a ferromagnet at every point.
6. Would the maximum possible magnetisation of a paramagnetic sample be of the same order of magnitude as the magnetisation of a ferromagnet?

Answer:
1. The tendency to disrupt the alignment of dipoles (with the magnetising field) arising from random thermal motion is reduced at lower temperatures.

2. The induced dipole moment in a diamagnetic sample is always opposite to the magnetising field, no matter what the internal motion of the atom is.

3. Slightly less, since bismuth is diamagnetic.

4. No, as is evident from the magnetisation curve. From the slope of the magnetisation curve, it is clear that μ is greater for lower fields.

5. Proof of the important fact (of much practical use) is based on boundary conditions of magnetic fields (B & H) at the interface of two media. (When one of the media has μ>>1, the field lines meet this medium nearly normally).

6. Yes. Apart from minor differences in the strength of the individual atomic dipoles of two different materials, a paramagnetic sample with saturated magnetisation will have the same order of magnetisation. But of course, saturation requires impractically high magnetising fields.

Plus Two Physics Magnetism and Matter One Mark Questions and Answers

Question 1.
Nickel shows ferromagnetic property at room temperature. If the temperature is increased beyond Curie temperature, then it will show
(a) anti ferromagnetism
(b) no magnetic property
(c) diamagnetism
(d) paramagnetism
Answer:
(d) paramagnetism

Question 2.
According to Curie’s law, the magnetic susceptibility of a substance at an absolute temperature T is proportional to
(a) 1/T
(b) T
(c) 1/T²
(d) T²
Answer:
(a) 1/T
Explanation : According to Curie’s law X ∝ \(\frac{1}{T}\)

Question 3.
Above Curie temperature
(a) a paramagnetic substance becomes ferromagnetic substance
(b) a ferromagnetic substance becomes paramagnetic
(c) a paramagnetic substance becomes diamagnetic
(d) a diamagnetic substance becomes paramagnetic
Answer:
(b) a ferromagnetic substance becomes paramagnetic

Plus Two Physics Magnetism and Matter Two Mark Questions and Answers

Question 1.
Classify the following properties into diamagnetic, paramagnetic and ferromagnetic.

1. Susceptibility -1≤x<0
2. In uniform magnetic field it acquires a large magnetisation in the direction of the field.
3. When it is suspended in a magnetic field, it will come to rest perpendiculartothe magnetic field.
4. Susceptibility of the substance varies inversely as temperature of the substance upto curie temperature ie. x_m∝ \(\frac{1}{T}\)

Answer:

• Dia – a,c
• Para – d
• Ferro – b

Question 2.
Classify the following properties into diamagnetism and ferro magnetism.

1. In non uniform magnetic field, it more from high to law field.
2. Magnetic field lines are repelled from this material, If we place in a external magnetic field.
3. susceptibility greater than one, <0, +ve
4. Iron, Nickel, Cobalt are the examples

Answer:

1. Diamagnetism
2. Diamagnetism
3. Ferro magnetism
4. Ferro magnetism

Question 3.
Fill in the blanks.

Answer:

1. negative
2. Weak magnetic field to strong magnetic field.
3. Individual atoms have tiny magnetic moments
4. µ_r >1

Plus Two Physics Magnetism and Matter Three Mark Questions and Answers

Question 1.
The figure shows hysteresis curves for soft iron and stell.

1. Which among the two is the hysteresis loop of soft iron?
2. Which one among the two materials is preferred for use in transformers and galvanometers?
3. When steel is once magnetized, the magnetiza-tion is not easily destroyed even if it is exposed to strong reverse fields. Give reason.

Answer:

1. Fig. b
2. Soft iron
3. When steel is magnetised the magnetic domains in the material is permanently set and magnetised permanently in the direction of the applied magnetic field.

Question 2.
The figure shows a liquid placed on the pole pieces of two magnets.

1. Which magnetic behaviour is exhibited by the liquid? (1)
2. Write any two characteristics of this magnetic behaviour? (1)
3. Does this behaviour transform with temperature. Why? (1)

Answer:
1. Diamagnetism
2. characteristics of this magnetic behaviour:

• Permeability of a diamagnetic material is less than one.
• Susceptibility is small and negative.

3. No. The magnetic dipole moment induced in the diamagnetic material is opposite to the magnetising field and hence does not affect the thermal motion of atoms. Hence change in temperature has no effect on diamagnetism.

Question 3.
A magnetic material contained in a curved glass plate, when placed in a nonuniform magnetic field, exhibits a property as shown in figure.

1. Which type of magnetic material is this? Explain the property
2. Write two examples for such a magnetic material. Explain how the property relates with temperature?
3. “The susceptibility of a material is small” what do you mean by this statement.

Answer:

1. Diamagnetism: Diamagnetic materials are repelled from external magnetic field.
2. Bismuth, Copper, Lead, Nitrogen, Water. Diamagnetism is independent of temperature.
3. Diamagnetism developed by external magnetic field is very small.

Question 4.
A place where dip = 90°, B_H = 0

1. This place is at…………
2. What is the value of B_u at this place?
3. Can a magnetic needle align in the N-S direction at this place?

Answer:
1. Magnetic pole.

2. B = \(\sqrt{B_{v}^{2}+B_{h}^{2}}\)
But B_h = 0
∴ B = B_v

3. No. Magnetic needle align vertically

Question 5.
Classify into diamagnet, Paramagnet and Fero magnet.

1. Feebly magnetized in the same direction
2. m_r slightly more than none
3. does not allow lines of force
4. temperature independent
5. exhibit hysteresis
6. strongly attracted by a bar magnet

Answer:

1. Paramagnet
2. Para magnet
3. Dia magnet
4. Diamagnet
5. Ferromagnet
6. Ferromagnet

Plus Two Physics Magnetism and Matter Four Mark Questions and Answers

Question 1.
Magnetic field lines are the visualization of magnetic field.

1. Write any three properties of magnetic field lines.
2. The arrangement shows two bar magnets placed near each other. Draw the magnetic field lines of the system.

Answer:
1. properties of magnetic field lines:

• The magnetic field lines of a magnet form continuous closed loops.
• The tangent to the field line at a given point represents the direction of magnetic field at that point.
• Flux density of magnetic field represents the strength of magnetic field.
• The magnetic field lines do not intersect

2.

Question 2.
The figure shows the magnetic field of earth.

1. Identify the labels A, B, C. (1)
2. The lines drawn on a map through places that have the same declination are called………(1)
3. The horizontal component of earth’s magnetic field at a place is 0.25 × 10^-4T and the resultant magnetic field is 0.5 × 10^-4 T. Find the dip and the vertical component of the earth’s magnetic field at the place. (2)

Answer:
1. the labels A, B, C:

• A – Magnetic equator
• B – Magnetic axis
• C – Declination

2. Isogonic lines

3. Horizontal component of earth’s magnetic field is
B_H = B cosδ
0.25 × 10^-4 = 0.5 × 10^-4 δ
δ = 60°
Vertical component of earth’s magnetic field is
Bv = B sin δ = 0.5 × 10^-4 sin 60 = 0.43 × 10^-4T.

Plus Two Physics Magnetism and Matter Five Mark Questions and Answers

Question 1.
When a magnetic needle is placed in a non-uniform magnetic field it experiences
1.

• a force but no torque
• a torque but no force
• force and torque
• neither a force nor a torque (1)

2. A bar magnet held perpendicular to a uniform magnetic field as in the figure. If the torque acting on it is to be reduced to 1/4^th by rotating the magnet towards the direction of the field, find the angle through which the magnet is to be rotated. (2)

3. State whether the potential energy of the magnet increases or decreases after rotation. Justify your answer. (2)
Answer:
1. force and torque

2. When the bar magnet is perpendicular to the field Torque is maximum
τ = MBsinθ = MBsin(90) = MB
When rotated through an angle θ, torque is

Angle through which the magnet is to be rotated is 90 – θ = 75.53°

3. Potential energy decreases. Potential energy is minimum when the magnet is parallel to the field. U = MBcosθ. When rotated to 0° (to make magnet parallel to the field) Potential energy decreases.

Students can Download Chapter 4 Planning Questions and Answers, Plus Two Business Studies Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Business Studies Chapter Wise Questions and Answers Chapter 4 Planning

Plus Two Business Studies Planning One Mark Questions and Answers

Question 1.
Planning function is required to be performed by all managers at all levels. This feature of planning is
Answer:
Planning is pervasive.

Question 2.
When planning is reduced to black and white, it is known as
Answer:
Formal planning.

Question 3.
Operational plan is undertaken at
Answer:
Lower level management.

Question 4.
Plans are intended to be used repeatedly
Answer:
Standing Plans.

Question 5.
Is a plan which states the expected results of a given future period in numerical terms.
Answer:
Budget.

Question 6.
Which function of management is considered as the base of all other functions?
Answer:
Planning.

Question 7.
Identify the management function which insists on ‘thinking before doing’.
Answer:
Planning.

Question 8.
“These are assumptions on the basis of which plans are formed”. Find the appropriate term used to denote this.
Answer:
Planning premises.

Question 9.
I am a general statement and i provide guidelines in decision making to members of an organization. Who am i?
(a) Procedure
(b) Rule
(c) Policies
(d) Project
Answer:
(c) Policies

Question 10.
‘Mobile phones are restricted in schools’. What type of plan is this? (Method, Programme, Rule, Policy)
Answer:
Rule.

Question 11.
Find out the correct pair.
(a) Objective – Promotion is based on merit only
(b) Policy – Time rate system of wage payment
(c) Rule – No smoking
(d) Method – Make a profit of 40% on capital invested
Answer:
(c) Rule – No smoking

Question 12.
The management of Arun Industries decided to implement the following with effect from the year2007 – “Every employees should mark their attendance before 8 am”. Which aspect of planning is referred here?
Answer:
Rule.

Question 13.
State the type of plan involved in the following two sentences.

1. Promotion is based on merit only
2. Smoking is prohibited in the factory premises.

Answer:

1. Policy
2. Rule

Question 14.
“We sell goods only on cash basis.” Which type of plan is mentioned here?
Answer:
Policy.

Question 15.
Select an example each of objecvitves, goals and policies from the following plans.

1. Increase sales by 10%
2. Every employee should mark attendance before 10 a.m.
3. Maximise the profit.

Answer:

1. Objective
2. Rule
3. Objective

Question 16.
‘Offering 30% of jobs to women.’ What type of plan is it?
Answer:
Policy.

Question 17.
Name the type of plan which provides a basis for interpreting the strategy.
Answer:
Policies.

Question 18.
Name the type of plan in which the minutest details are worked out, that is procedures, rule and budget within the broad framework of policy.
Answer:
Programme.

Question 19.
Name the type of plan which is also a control device from which deviation can be taken care of
Answer:
Budget.

Question 20.
Sony electronics decides to sell television sets through exchange of old television sets. What type of plan is it?
Answer:
Strategy.

Question 21.
Which plan suggests acts and non-action of employees?
Answer:
Rules.

Question 22.
Detect the differences in the following statements contained in a firm’s plan.

1. We sell goods only on Cash basis
2. Smoking is prohibited in the factory Premises.

Answer:

1. Policy
2. Rule

Question 23.
Which among the following is not a limitation of planning?
(a) It reduces flexibility
(b) It is futuristic
(c) Does not guarantee success
(d) Planning is pervasive
i) a&b ii) b&d iii) a, b&d iv)c
Answer:
(ii) b&d

Question 24.
Match the following.

Answer:
a – ii
b – v
c – i
d – iii

Question 25.
Classify the following as single use plan and standing plan.
Budget, Method, Rule, Programmes, Procedure & Policies
Answer:
1. Single use Plan
Budget, Programme.
2. Standing Plan
Method, Rule, Procedure, Policies.

Plus Two Business Studies Planning Two Mark Questions and Answers

Question 1.
What is planning?
Answer:
Planning – Meaning:
Planning is deciding in advance what to do and how to do. It is one of the basic managerial functions. Planning is closely connected with creativity and innovation. Planning involves setting objectives and developing appropriate courses of action to achieve these objectives.

Question 2.
State any two consequences if there was no planning.
Answer:

1. If there was no planning employees would be working in different directions.
2. The management would not be able to achieve.

Plus Two Business Studies Planning Three Mark Questions and Answers

Question 1.
In a classroom discussion, Kalesh argued that planning is a function to be performed only at the top-level management. But Suresh is of the opinion that planning is required at all levels of management.

1. Do you support the argument of Kalesh or Suresh?
2. Justify your answer

Answer:
1. support the argument of Suresh
2.Planning function is required to be performed by the managers at all levels. Board of Directors plan at top level, Functional managers at departmental level and foreman at lower level. The degree and importance of planning depend on the level at which it is undertaken.

Question 2.
“Planning is needed in every function and at every level”. Explain.
Answer:
Planning function is required to be performed by the managers at all levels. Board of Directors plan at top level, Functional managers at departmental level and foreman at plant level. The degree and importance of planning depend on the level at which it is undertaken.

Question 3.
Planning is not a guarantee of success of a business. Comment.
Answer:
It is right to say that planning is not a guarantee of success of business. Since it based on assumptions regarding future and assumptions cannot be hundred percent accurate.

Question 4.
How does planning help in co-ordination?
Answer:
Planning makes co-ordination of various activities, departments and groups relatively easy. Co-ordination of departmental operations is facilitated by planning through the establishment of common goals.

Plus Two Business Studies Planning Four Mark Questions and Answers

Question 1.
“Planning is a continuous process”.

1. Do you agree?
2. Explain

Answer:
1. Yes.

2. Planning is an ongoing and dynamic process. Business environment undergoes constant changes and accordingly, plans have to be modified. The completion of one plan requires the other plan to be undertaken. Hence planning is a never-ending activity.

Question 2.
Match the following.

Answer:

Question 3.
What do you mean by “Single Use Plan” and “Standing Plan”?
Answer:
Types of Plans
1. Single-use plan:
A single-use plan is developed for a one-time event or project. Such plans are not to be repeated in future. E.g. budgets, programmes, projects, etc.

2. Standing plan:
A standing plan is used for activities that occur regularly over a period of time. E .g. policies, procedures, methods and rules.

Plus Two Business Studies Planning Five Mark Questions and Answers

Question 1.
“Forecasting is the essence of planning.”

1. Do you agree?
2. Explain.

Answer:
1. Yes.

2. Forecasting is the essence of planning. Forecasting involves assessing the future. Forecasting is a process of predicting relevant future situations that are likely to affect the activities of the organisation. Customer’s demand, competition, government policies, etc. can be assessed through forecasting. Forecasting helps an organisation to prepare plan efficiently and effectively.

Question 2.

1. Explain the following diagram and identify the management function.
2. Explain its features.

Answer:
1. management function.

2. Features of planning:

• Planning is goal oriented
• It is the primary function of management
• It is required at all levels of management
• Planning is a continuous process
• Planning is futuristic (forward looking)
• It is a decision making function
• It is a mental process

Question 3.
The Board of Directors of Karunya Ltd. decided to implement the following plans in their personnel department during the year 2006.

1. Filling up the vacancies through promotion
2. Selection of workers through various tests such as aptitude test, trade test, interview, etc.
3. Newly selected workers should be given two weeks apprenticeship training.
4. All employees should mark their attendance before 8 am.
5. Increase production of a product by 10% in the year 2018-19.

You are required to

• Classify the plans adopted by Karunya Ltd.
• Establish the reasons for such classification.

Answer:

1. Policy – Guide in decision making
2. Procedure – Manner in which activities are performed
3. Programmes – All activities necessary for achieving a given objective
4. Rules – Rule is a rigid policy which cannot be altered.
5. Objective – Target to be achieved.

Question 4.
“Planning, the primary function of management, suffers from a number of limitations.” Elucidate.
Answer:
Limitations of Planning

1. Planning makes the activities rigid.
2. Long term plans are insignificant in the rapidly changing business environment.
3. It reduces creativity.
4. It involves cost.
5. It involves a lot of time.
6. Planning does not guarantee success.

Plus Two Business Studies Planning Eight Mark Questions and Answers

Question 1.
“No enterprise can achieve its objectives without systematic planning.”

1. Do you agree with this statement?
2. Give reasons in support of your answer.

Answer:
Importance of Planning
1. Planning provides directions:
By stating in advance how work is to be done planning provides direction for all actions.

2. Planning reduces the risk of uncertainty:
Planning enables an organisation to predict future events and prepare to face unexpected events.

3. Planning reduces wasteful activities:
Planning serves as the basis of co-ordinating the activities and efforts of different departments and individuals. It helps to eliminate useless and redundant activities.

4. Planning promotes innovative ideas:
Since planning is thinking in advance, there is scope for finding better and different methods to achieve the desired objectives.

5. Planning facilitates decision making:
Planning helps in decision making by selecting the best alternative among the various alternatives.

6. Facilitates control:
Planning provides the basis for control. Planning specifies the standard with which the actual performance is compared to find out deviation and taking corrective action.

Question 2.
Briefly explain the steps in planning.
Answer:
Planning Process (Steps in Planning)
1. Setting Objectives:
The first step in planning

2. Developing premises:
Planning is based on certain assumptions about the future. These assumptions are called planning premises. Forecasting is important in developing planning premises.

3. Identifying alternative courses of action:
The next step in planning is to identify the alternative courses of action to achieve the objectives.

4. Evaluating alternative Courses:
The pros and cons of various alternatives must be evaluated in terms of their expected cost and benefits.

5. Selecting an alternative:
After evaluating the alternatives the manager will select that alternative which gives maximum benefit at minimum cost.

6. Implement the plan:
Implementation of a plan means putting plans, into action so as to achieve the objectives of the business.

7. Follow up action:
Plans are to be evaluated regularly to check whether they are being implemented and activities are performed according to schedule.